Solutions and Colligative Properties

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2 46 Objective MHT-CET Chemistry 2 Solutions and Colligative Properties 2.1 Introduction 2.2 Types of Solutions 2.3 Concentration of Solutions of Solids in Liquids 2.4 Solubility of Gases in Liquids 2.5 Solid Solutions 2.6 Colligative Properties 2.7 Lowering of Vapour Pressure 2.8 Boiling Point Elevation 2.9 Freezing Point Depression 2.10 Osmosis and Osmotic Pressure 2.11 Abnormal Molecular Masses 2.12 van't Hoff Factor 2.1 Introduction Two-third of the earth is covered with water, the most essential, vital component for life. Water in the sea is not in pure form. It contains sodium chloride, magnesium chloride, calcium salts and gases like oxygen, carbon dioxide in dissolved state. Thus, sea water is a solution. Solution consists of at least two components, a medium and dissolved phase. The physical properties of solution are mainly originated by intermolecular forces of attraction between the solvent molecules. These intermolecular forces of attraction are changed by the presence of dissolved substances. The physical properties include vapour pressure, boiling point, freezing point and osmotic pressure. Solutions Solutions are mixtures of two or more components. Depending on sizes of the components, the mixtures are classified into three types : A coarse mixture : It is formed when the sizes of the constituent components are relatively bigger, e.g.; mixture of salt and sugar. A colloidal dispersion : It is formed when the sizes of the particles dispersed in solvent are in the range of 10 7 cm to 10 4 cm. Colloidal particles carry positive or negative charge which stabilizes colloidal dispersion e.g.; ferric hydroxide sol, arsenic sulphide sol, etc. Colloidal solutions are heterogeneous and can be easily separated. A true solution : A true solution is formed when soluble substances are dissolved in the solvent. The sizes of the particles dissolved are very small of the order of 10 8 cm. True solutions are homogeneous and cannot be separated into components by simple mechanical methods. ff A true solution is defined as a homogeneous mixture of two or more substances, the composition of which is not fixed and may be varied within certain limits. ff A solution is formed by two components, solvent and one or more solutes. The component of the solution which constitutes larger part of the solution is called solvent and the other component that constitutes smaller part is called solute. Homogeneous solution : Solution is homogeneous if its composition is uniform throughout the body of the solution. Heterogeneous solution : Solution is heterogeneous when two or more phases are present in it. Solvation : The process of interaction of solvent molecules with solute particles to form aggregates. When water is a solvent, the process of solvation is called hydration or aquation. ff The extent to which solute dissolves in solvent to form homogeneous solution depends on nature of solute and solvent. ff The general rule is, like dissolves like i.e., polar

3 Solutions and Colligative Properties solutes are soluble in polar solvents, (e.g.; NaCl in water) while non-polar solutes are soluble in nonpolar solvents (e.g.; iodine in CCl 4 ). Water is called universal solvent, as it is polar and has very high dielectric constant, hence, dissolves most of the polar solutes. Solutions containing two, three or four components are called binary, ternary and quaternary solutions respectively. Solutions prepared in water are called aqueous solutions and solutions in other solvents are called non-aqueous solutions. As the solute particles are very small, the components of true solutions cannot be separated by simple physical methods like centrifugation, filtration, etc. 2.2 Types of Solutions Solvent Solute Examples Solid Solid Alloys like brass, bronze, copper in gold etc. Solid Liquid Amalgams of mercury with metals 47 Solid Gas Hydrogen gas in palladium metal, pumice stone Liquid Solid Iodine in CCl 4, benzoic acid in C 6 H 6, sugar in water Liquid Liquid Ethanol in water Liquid Gas Oxygen, carbon dioxide in water Gas Solid Iodine in air Gas Liquid Chloroform in nitrogen Gas Gas Air, mixtures of non-reacting gases 2.3 Concentration of Solutions of Solids in Liquids The concentration of a solution is defined as the amount of solute dissolved in a specific amount of solvent. Solutions containing relatively less amount of solute are called dilute solutions and if it contains relatively more amount of solute then the solution is called concentrated solution. Concentration of solutions may be expressed in different ways as discussed below : Name Symbol Formula Definition Percentage by mass Volume percentage Mass by volume percentage Strength Parts per million %(w/w) %(v/v) Mass of solute Total mass of solution 100 Amount of solute in grams present in 100 g of solution. Volume of solute Total volume of solution 100 Volume of solute in ml dissolved in 100 ml of the solution. % (w/v) Mass of solute Total volume of solution in ml 100 Amount of solute in grams dissolved in 100 ml of the solution. g/l (or g/dm 3 ) ppm Mass of solute in grams 3 Volume of solution in L (or dm ) Mass or volume of solute Total mass or volume of solution 106 Molarity M Moles of solute Volume of solution in L(or dm 3 ) Molality m Moles of solute Mass of solvent in kg Mole fraction x n xa A na + nb Amount of solute in grams present in one litre (or dm 3 ) of solution. Number of parts of solute present in million (10 6 ) parts of solution. Number of moles of solute dissolved in one litre (or one dm 3 ) of solution. Number of moles of solute dissolved in 1 kg of the solvent. Ratio of number of moles of one component to the total number of moles of all the components. Effect of temperature No effect Changes with change of temperature. Changes with change of temperature. Changes with change of temperature. No effect Changes with change of temperature. No effect No effect

4 48 Objective MHT-CET Chemistry Illustration : 1.23 g of sodium hydroxide (molar mass 40) are dissolved in water and the solution is made to 100 cm 3. Calculate the molarity of the solution. Soln.: Amount of NaOH 1.23 g Volume of solution 100 cm 3 Mass of NaOH Moles of NaOH Molar mass of NaOH Moles of NaOH Molarity 1000 Volume of solution M 100 Illustration : 1.8 g of glucose (molar mass 180) are dissolved in 60 g of water. Calculate (a) the molality (b) mole fraction of glucose and water. Soln.: (a) Molality of solution Moles of solute Mass of solvent in g 1000 Moles of glucose \ Molality 1000 Mass of water m (b) Mole fraction Moles of solute Moles of solute + Moles of solvent 18. Moles of glucose 001. ; Moles of water Mole fraction of glucose ; Mole fraction of water Illustration : 4 g of sodium chloride was dissolved in 300 g of water. Calculate percentage by mass of sodium chloride in solution. Soln.: Percentage by mass of sodium chloride (w/w) mass of sodium chloride mass of sodium chloride + mass of water g %by mass 4 g+ 300g 304 Depending upon the quantity of solute dissolved in a liquid solvent, solutions can be of three types : f f Saturated solution : A solution which cannot dissolve any further amount of solute at a given temperature is called saturated solution. f f Unsaturated solution : A solution in which more amount of the solute can be dissolved at a given temperature is called unsaturated solution. f f Supersaturated solution : A solution in which amount of solute is more than it can dissolve at a particular temperature is called supersaturated solution. Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. Factors affecting solubility of a solid in a liquid : f f Effect of temperature : If the dissolution process is endothermic (D sol H > 0), the solubility increases with rise in temperature. If dissolution process is exothermic (D sol H < 0) the solubility decreases with rise in temperature. f f Effect of pressure : Pressure does not have any significant effect on solubility of solids in liquids as these are highly incompressible. 2.4 Solubility of Gases in Liquids Gases are soluble in liquids including water. The solubility of gases like O 2, N 2, etc. are much low. O 2 molecules being non-polar, have less solubility in polar solvent, water. CO 2 and NH 3 gases are more soluble in water as CO 2 reacts with water to form carbonic acid and NH 3 reacts with water to form ammonium hydroxide. HCl gas is polar, its solubility is very high in water, forming hydrochloric acid. Factors affecting solubility of a gas in a liquid : f f Effect of pressure : The solubility of gas increases with the increase in external pressure. Henry s law - It states that the solubility of a gas in a liquid at constant temperature is proportional to the pressure of the gas above the solution. S P i.e., S K P where S is the solubility of the gas in mol dm 3, P is the pressure of the gas in atm, K is constant of proportionality and has the unit of mol dm 3 atm 1. If P 1 atm, then S K. Hence, Henry s constant K is defined as solubility of gas in mol dm 3 at 1 atmospheric pressure at reference temperature. If several gases are present then the solubility of any gas may be evaluated by using P as partial pressure of that gas in the mixture. ff Effect of temperature : According to Charles law, volume of a given mass of a gas increases with increase of temperature. Therefore, volume

5 Solutions and Colligative Properties of a given mass of dissolved gas in solution also increases with increase of temperature, so that it becomes impossible for the solvent in solution to accommodate gaseous solute in it and gas bubbles out. Hence, solubility of gas in liquid decreases with increase of temperature. f f Effect of addition of soluble salt : Solubility of dissolved gas is suppressed when a soluble salt is added to the solution of gas. Illustration : The solubility of nitrogen gas at 1 atm pressure at 25 C is mol dm 3. Calculate the solubility of N 2 gas from atmosphere at 25 C if atmospheric pressure is 1 atmosphere and partial pressure of N 2 gas at this temperature and pressure is 0.78 atm. Soln.: (i) S mol dm 3, P 1 atm N2 S K P N mol dm 3 K 1 atm K mol dm 3 atm 1 (ii) P 0.78 atm, S? S mol dm 3 atm atm mol dm 3 i.e. Solubility of N 2 is reduced to mol dm Solid Solutions A solid solution of two or more metals or of a metal or metals with one or more non-metals is called an alloy or solid solution. All the properties of the pure metals are improved when they form solid solutions, i.e., alloys. Duralumin (Al + Cu + Mg + Mn) : Light and strong as steel. Used in the construction of aircrafts. Aluminium bronze (Al + Cu + Mn) Lead alloy (Pb % Sb) : Acid resistant. Used for bearings, bullets, shrapnel and for manufacturing lead storage battery plates. Babbitt metal (Sb + Sn + Cu) : Antifriction alloy. Used in machine bearings. Stainless steel (Steel + Cr + Ni) : Resistant to corrosion. Used in cutlery. 49 Spiegeleisen (5-20% Mn in Iron ) and Ferromanganeous (70-80% Mn % Fe) : Used for making very hard steels and to manufacture rails, safes and heavy machinery. Manganin (84% Cu + 12% Mn + 4% Ni) : Has almost zero temperature coefficient of electrical resistance. Used for making electrical measurements. Amalgams (Hg + metals) : Used for extracting metals from ores. 2.6 Colligative Properties These are the properties that depend on the number of solute particles in solution and not on the nature of the solute particles. These are : Lowering of vapour pressure, Elevation of boiling point of solvent in solution, Depression of freezing point of solvent in solution and Osmotic pressure. 2.7 Lowering of Vapour Pressure Vapour pressure of liquids : ff The vapour pressure of a substance is defined as the pressure exerted by the gaseous state of that substance when it is in equilibrium with the solid or liquid phase. ff Vapour pressure of a liquid, increases with the increase of temperature. ff The boiling point of a liquid is a temperature at which vapour pressure of liquid becomes equal to external pressure. If the boiling is carried out in an open atmosphere then external pressure is the atmospheric pressure. Vapour pressure lowering : ff The vapour pressure of a liquid solvent is lowered when a non-volatile solute is dissolved in it to form a solution. ff This is due to the fact that in case of pure solvent, its surface area is completely occupied by volatile solvent molecules. While in case of solution of nonvolatile solute, its surface area is not completely available for volatile solvent; partly it is occupied by non-volatile solute. ff Hence, rate of evaporation of the solution will be less as compared to that of pure solvent and vapour pressure of solution is lower than that of the pure solvent. ff If p 1 is the vapour pressure of pure solvent and p is the vapour pressure of the solution of non-volatile

6 50 Objective MHT-CET Chemistry ff ff solute in the same solvent, then p < p 1 and the lowering of vapour pressure is, Dp p 1 p The difference between vapour pressure of pure solvent and the vapour pressure of solvent from solution is called vapour pressure lowering. The ratio of vapour pressure lowering of solvent from solution to the vapour pressure of pure solvent is called the relative lowering of vapour pressure. Dp p p p p 1 Raoult s law 1 1 The law states that, the partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution. p 1 p 1 x 1 and p 2 p 2 x 2 where p 1 and p 2 are vapour pressures of pure components 1 and 2 respectively, at the same temperature. Total vapour pressure, p total of solutions of two volatile components is the sum of partial vapour pressures of the two components, p total p 1 + p 2 x 1 p 1 + x 2 p 2 (x 1 )p 1 + (1 x 1 ) p 2 p 2 + (p 1 p 2 )x 1 The solution which obeys Raoult s law over the entire range of concentration is called an ideal solution. If a solution does not obey Raoult s law, the solution is nonideal. If y 1 and y 2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, p 1 y 1 p total and p 2 y 2 p total Raoult s law for a solution of non-volatile solute Non-volatile solute does not evaporate and does not contribute to the total vapour pressure of solution. The vapour pressure of pure component A 1 is p 1 and that of component A 2 is p 2 0. p p 2 + (p 1 p 2 )x 1 (as p 2 0) p 0 + (p 1 0)x 1 i.e., p p 1 x 1 (1) The lowering of vapour pressure Dp is given by, Dp p 1 p p 1 p 1 x 1 p 1 (1 x 1 ) But 1 x 1 x 2 Hence, Dp p 1 x 2 (2) Lowering of vapour pressure is the product of vapour pressure of pure solvent and mole fraction of nonvolatile solute dissolved in volatile solvent to form a solution. Thus the lowering of vapour pressure depends on nature of pure solvent and concentration of solute in mole fraction. Now the relative lowering of vapour pressure is given by, Dp p 1 p px 1 2 x2 p 1 p 1 p 1 Hence, relative lowering of vapour pressure x 2 Equation (2) proves that the lowering of vapour pressure is a colligative property because it depends on the concentration of non-volatile solute. Molar mass of solute and relative lowering of vapour pressure Dp p 1 p n2 W2 / M 2 x2 p p n n W M W M / 1+ 2 / 2 For dilute solutions n 1 >> n 2 \ Dp n p 2 W M WM n 2 / W M WM 1 1 1/ Knowing the masses of non-volatile solute and the solvent in dilute solutions and by determining experimentally vapour pressure of pure solvent and the solution it is possible to determine molar mass of a non-volatile solute. Illustration : Calculate the weight of a non-volatile solute (mol. wt. 40), which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. Soln.: Given: p (80/100)p, M 2 40 W g, M We know, p p W2 M1 p M2 W1 p ( / ) p W \ p \ W 2 18 g (weight of solute required) 2.8 Boiling Point Elevation Boiling point is defined as the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure. ff It is a characteristic property of liquids and is a criterion to check the purity of liquid. ff It increases with increase in external pressure. ff Liquids having greater intermolecular forces have high boiling points.

7 Solutions and Colligative Properties Elevation of boiling point : ff ff Solution has lower vapour pressure and hence higher boiling point than pure solvent. The increase in the boiling point, DT b T b T b is known as elevation of boiling point. For dilute solutions, DT b m or DT b K b m K W B 1000 b MB WA 1000 W K or M B b B DTb WA where m is molality of solution and K b is called boiling point elevation constant or molal elevation constant or ebullioscopic constant, having unit K kg mol 1. Illustration : The molal elevation constant for water is 0.51 K kg mol 1. Calculate the b. pt. of solution made by dissolving 6 g urea in 200 g water. Soln.: Given: W 2 6 g, M 2 60 (urea), W g, K b 0.51 K kg mol 1 We know, DT b 1000 Kb W2 M2 W Substituting values, DT b C As elevation in b. pt C \ B. pt. of solution C 2.9 Freezing Point Depression Freezing point of a liquid is the temperature at which the vapour pressure of solid is equal to the vapour pressure of liquid. Depression of freezing point : The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent. The decrease in freezing point, DT f T f T f is known as depression in freezing point. Freezing point depression and vapour pressure lowering : For dilute solutions, DT f p 1 p DT f m or DT f K f m where K f is known as freezing point depression constant or molal depression constant or cryoscopic constant, having unit K kg mol Depression in freezing point and molar mass of the solute : DT f K W B 1000 f M W or M K f W B 1000 B B A DTf WA Illustration : The freezing point of a solution containing 50 cm 3 of ethylene glycol in 50 g water is found to be 34 C. Assuming ideal behaviour, calculate the density of ethylene glycol. (K f for H 2 O 1.86 K kg mol 1 ) Soln.: Given: V ethylene glycol 50 cm 3, W 1 50 g K 1.86 K kg f(h2 O) mol 1, M 2 62 (glycol) DT f 34 C, W 2 50 d (glycol) We know, DT f 1000 Kf W2 M2 W ( 50 d) Substituting values, \ d g/cm Osmosis and Osmotic Pressure Osmosis : The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane. Osmotic pressure : The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane. ff Two or more solutions having same osmotic pressure at a given temperature are called isotonic solutions. ff If one solution is of lower osmotic pressure, it is called hypotonic with respect to the more concentrated solution. The more concentrated solution is said to be hypertonic with respect to the dilute solution. ff If a pressure higher than the osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through the semipermeable membrane and the process is called reverse osmosis. It is used in desalination of sea water. Laws of osmotic pressure van t Hoff Boyle s law : It states that, at constant temperature the osmotic pressure (p) of a dilute solution is directly proportional to its molar concentration or inversely proportional to the volume of the solution. At constant T, p C (in mol L 1 )

8 52 Objective MHT-CET Chemistry If n moles of solute is dissolved in V litres then, C n V n 1 \ p or p for constant n V V \ pv constant or p C constant van t Hoff-Charles law : It states that, the concentration remaining constant, the osmotic pressure of a dilute solution is directly proportional to the absolute temperature. At constant C, p T i.e., p T constant van t Hoff general solution equation : p CRT R J K 1 mol 1 if p is in N m 2 and V in m 3 R L atm K 1 mol 1 if p is in atm and V in dm 3 Determination of molar mass from osmotic pressure : p n 2 V RT, p W 2 RT, M 2 W 2 RT M2 V p V van t Hoff Avogadro s law : It states that, two solutions of equal concentrations of different solutes exert same osmotic pressure at the same temperature. It can also be, stated as, equal volumes of isotonic solutions contain an equal number of solute particles at the given temperature. For a given solution pv nrt p 1 V 1 n 1 RT 1 for solution 1 (i) p 2 V 2 n 2 RT 2 for solution 2 (ii) If p 1 p 2, T 1 T 2 and V 1 V 2 then from equations (i) and (ii), n 1 n 2 Since, number of moles are equal, number of molecules are also equal. Hence, osmotic pressure and temperature remaining the same, equal volumes of solutions would contain equal number of moles of the solute. For isotonic solutions of equal volume, n 1 n Abnormal Molecular Masses For substances undergoing association or dissociation in solution, the molecular mass determined by studying any of the colligative properties is different than the theoretically expected value, and the substance is said to show abnormal molecular mass. Dissociation of electrolyte solutes : Electrolytic solutes when dissolved in solvent dissociate to produce multiple number of ions/particles. This results in increase of number of solute particles in solution which in turn results in increase of colligative properties and increase of DT f /m value and the value of DT f /m is approximately equal to integral multiple of K f value. The value of integral is equal to total number of ions produced on dissociation as shown : (i) HCl H + + Cl ; 2 particles; DT f /m K mol 1 kg + (ii) NH 4 Cl NH 4 + Cl ; 2 particles; DT f /m K mol 1 kg (iii) CoCl 2 Co Cl ; 3 particles; DT f /m K mol 1 kg (iv) K 2 SO 4 2K + + SO 2 4 ; 3 particles; DT f /m K mol 1 kg (v) AlCl 3 Al Cl ; 4 particles; DT f /m K mol 1 kg DT f /m K f value observed in case of solutions of electrolytes may not be exactly two fold, three fold etc of theoretical K f value observed in case of solutions of non-electrolyte solute. The value fluctuates with degree of dissociation of solute in solution. Association of solutes : In some non-polar solvents, two or more molecules of solute associate to form bigger molecules. For example, in benzene solutes like acetic acid, benzoic acid, etc. associate to form dimers. This association is due to the hydrogen bonding between these molecules. 2CH 3 COOH (CH 3 COOH) 2 2C 6 H 5 COOH (C 6 H 5 COOH) 2 Hence, numbers of solute particles are reduced to almost half. Observed molecular masses of these species are almost twice the expected values in dilute solutions. Due to association the total number of molecules in solution will be almost half of the number of molecules of the substance dissolved. Hence such solutions show abnormally low colligative properties. The observed molecular masses are almost double van t Hoff Factor The colligative properties of electrolytes may be expressed in relation to colligative properties of nonelectrolytes by using van t Hoff factor i. It is defined as the ratio of observed colligative property produced by a given concentration of electrolyte solution to the property observed for the same concentration of nonelectrolyte solution.

9 Solutions and Colligative Properties Observed value of the colligative property i Theoretical value of the colligative property or i Theoretical molecular mass Observed molecular mass Mth Mo Total number of moles of particles after association/ dissociation i, Number of moles of particles before association / dissociation for association, i < 1; for dissociation, i > 1 van t Hoff factor ( i) and degree of dissociation (a) : i a 1 n 1 ; M(theoretical) i [ 1+ ( n 1) a] M(observed) M(theoretical) M(observed) a M(observed) ( n 1) van t Hoff factor ( i) and degree of association (a) : 1 i a 1 1/n Modified equations for colligative properties : p1 p1 i n 2, DT b ik b m, DT f ik f m, p i n 2 RT/V p1 n1 53 Illustration : A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate the osmotic pressure of the solution (R JK 1 mol 1 ). Soln.: Initial number of moles of K 4 [Fe(CN) 6 ] 0.1, (as the solution is decimolar). a degree of dissociation 50% 1/2 Thus, we may represent the facts as below : K 4 [Fe(CN) 6 ] 4K + + [Fe(CN) 6 ] 4 total Initial no of moles After 0.1(1 a) 0.1 (4a) 0.1 (a) 0.1(1 + 4a) dissociation 0.1 [1 + (4 1/2)] 0.3 van t Hoff factor, Number of particles after dissociation i Actual number of particles before dissociation pv inrt p inrt V icrt (C n/v molarity 0.1 mol L 1 ) or 01. mol 10 3 m mol m 3 or, p Nm atm

10 54 Objective MHT-CET Chemistry Mass Percentage : ( ww / )% of solution Mass of solute Total mass ofsolution 100 Volume Percentage : (/ vv )% of solution Volume of solute Total volume ofsolution 100 Mass by volume percentage : ( WV / )% of solution Mass of solute Total volume ofsolution inml 100 Moles Moles SOLUTIONS AND COLLIGATIVE PROPERTIES within Henry s Law : It states that the solubility of a gas in a liquid at constant temperature is proportional to the pressure of the gas above the solution. S P i.e., S K P i p A p A x A p B p B x B Isotonic Solutions : Solutions having same osmotic pressure. Hypotonic Solutions : Solutions having lower osmotic pressure than the other. Hypertonic Solutions : Solutions having higher osmotic pressure than the other. Laws of Osmotic Pressure: van t Hoff-Boyle s law : At constant T, V constant or 1 constant C ( in moll ) van t Hoff-Charles law : At constant C, constant van t Hoff generalsolutionequation : CRT van t Hoff-Avogadro s law : For isotonic solutions of equal volume, n1 n2. Strength T Mass of solute in grams 3 Volume of solution inl(or dm ) van t Hoff factor : i Observed valueof colligativeproperty Theoretical valueof colligativeproperty Theoretical valueofmolecularmass Observed valueofmolecularmass Degree of Dissociation andassociation : 1 (dissociation) i n 1 1 (association) i 1 1/ Modified Equations of Colligative Properties : p p in p n Tb ikm b Tf ikm f in RT / V 2 n x x K K K K Osmosis : The spontaneous and unidirectional flow of solvent molecules through a semipermeable membrane, into the solution or flow of solvent from a solution of lower concentration to the solution of higher concentration through a semipermeable membrane is called osmosis. The excess of pressure on the side of solution that stops the net flow of solvent into solution through semipermeable membrane is called osmotic pressure.

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12 Solutions and Colligative Properties 55 Multiple Choice Questions LEVEL Introduction 1. The mixture of salt and sugar is called a (a) coarse mixture (b) homogeneous mixture (c) racemic mixture (d) solution mixture. 2. In a solution, the larger proportion of the component is known as (a) solution (b) solute (c) solvent (d) mixed solution. 3. A solution is defined as a (a) homogeneous mixture of two or more substances (b) heterogeneous mixture of two or more substances (c) homogeneous mixture of liquid and solid components only (d) homogeneous mixture consisting water as one of the components. 4. The substances involved in solution are called (a) mole (b) atoms (c) molecules (d) components. 5. The scientist who won the first Noble prize in chemistry in the year 1901, for his work in solutions was (a) van t Hoff (b) Beckmann (c) Raoult (d) Jean Nollet. 6. In solution, the molecular size of particles is of the order of (a) 10 8 cm (b) 10 9 m (c) > 10 9 m (d) 10 9 cm 7. The new method to determine alcohol content of a wine was discovered in 1870 by (a) Newton (b) Joule (c) Raoult (d) C.A. Wurtz. 8. A solution having three components is called a (a) quaternary solution (b) binary solution (c) single solution (d) ternary solution. 2.2 Types of Solutions 9. An example of a solution having liquid in gas is (a) moist air (b) dry air (c) Au-Hg (d) C 2 H 5 OH + H 2 O 10. Which of the following is not a solution? (a) Air (b) A gold ring (c) Smoke (d) Salt solution 11. Solutions are of types. (a) three (b) six (c) nine (d) eleven 12. Which of the following possesses physical states of solute and solvent as liquid and solid respectively? (a) Solution of sugar in water (b) Zinc amalgam (c) Solution of naphthalene in benzene (d) Brass 13. Which of the following pairs of solutions having different physical states of solute? (a) Homogeneous mixture of chloroform in N 2 gas, solution of CO 2 in water (b) Brass, homogeneous mixture of camphor in N 2 gas (c) Sodium amalgam, moist air (d) Solution of H 2 gas in Pd metal, mixture of N 2 and O In vaporization of liquids into air, liquid dissolves into air hence liquid is (a) solution (b) solute (c) solvent (d) mixture. 15. Sugar dissolved in water is a type of (a) solid in solid solution (b) solid in gas solution (c) solid in liquid solution (d) gas in solid solution. 2.3 Concentration of Solutions of Solids in Liquids 16. Dissolution of calcium chloride in water is exothermic process while that of ammonium nitrate is endothermic process, the observation is

13 56 Objective MHT-CET Chemistry (a) (b) (c) (d) solubility of calcium chloride decreases with increase in temperature and that of ammonium nitrate increases with increase in temperature solubilities of both decrease with increase in temperature solubilities of both increase with increase in temperature solubility of calcium chloride increases with increase in temperature and that of ammonium nitrate decreases with increase in temperature cm 3 of C 2 H 5 OH on dissolving in 200 cm 3 of water forms 220 cm 3 solution of C 2 H 5 OH. What would be the percentage by volume of C 2 H 5 OH in the solution? (a) 22% (b) 11% (c) 01% (d) 10% 18. The volume of 2 M NaOH solution is 0.1 dm 3. What will be the volume of decimolar NaOH in dm 3? (a) 0.1 (b) 0.2 (c) 1.5 (d) A molal solution is one that contains one mole of a solute in (a) 1000 g of the solvent (b) one litre of the solvent (c) one litre of the solution (d) 1000 g of the solution. 20. When 10 g of caustic soda is dissolved in 250 cm 3 of water, molarity of the solution is (a) 1 M (b) 0.5 M (c) 0.25 M (d) 0.1 M 21. A solution contains 0.5 mole of a solute in 400 g of water, its molality would be (a) 1.25 m (b) m (c) 2.51 m (d) m 22. One part of solute in one million parts of solvent is expressed as (a) ppm (b) milligrams/100 cc (c) grams/litre (d) grams/100 cc g of glucose (molar mass 180 g/mol) is present in 500 g of water, the molarity of the solution is (a) 0.2 M (b) 0.4 M (c) 0.8 M (d) 1.0 M 24. The molality of 648 g of pure water is (a) 36 m (b) 55.5 m (c) 3.6 m (d) 5.55 m 25. An aqueous solution contains 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of acetic acid in this solution. (a) (b) (c) (d) When 0.5 g of solute is present in 10 7 g of a solution, the ppm of solute will be (a) 0.5 (b) 0.05 (c) 10 6 (d) g of urea was dissolved in 500 g of water. The percentage (by mass) of urea in the solution is (a) 0.86% (b) 1.186% (c) 11.86% (d) 0.08% cm 3 of ethyl alcohol was dissolved in 400 cm 3 of water to form 454 cm 3 of solution of ethyl alcohol. The percentage by volume of ethyl alcohol in water is (a) 0.12.% (b) 1.2% (c) 12.78% (d) 0.01% g of ethyl alcohol (molar mass 46 g mol 1 ) is dissolved in 54 g of water (molar mass 18 g mol 1 ). The mole fraction of ethyl alcohol and water in solution are respectively (a) 0.7 and 0.3 (b) and (c) and (d) and % (w/w) solution of ethylene glycol in water, an antifreezer is used in cars as coolant. It lowers freezing point of water to 17.6 C. The mole fraction of ethylene glycol is (a) (b) (c) (d) A solution of NaOH (molar mass 40 g mol 1 ) was prepared by dissolving 1.6 g of NaOH in 500 cm 3 of water. The molarity of solution is (a) 0.2 mol dm 3 (b) 0.02 mol dm 3 (c) 0.08 mol dm 3 (d) 0.8 mol dm g of urea was dissolved in 100 g of water. The molality of solution is (N 14, H 1, C 12, O 16) (a) mol kg 1 (b) mol kg 1 (c) mol g 1 (d) mol g g of sugar was dissolved in water to produce g of sugar syrup. The molality and mole fraction of sugar in the syrup are respectively (C 12, H 1, O 16) (a) m, (b) 0.05 m, 0.02 (c) m, (d) 0.05 m, The molarity and molality of sulphuric acid solution of density g cm 3 containing 27% by mass of sulphuric acid (molar mass of H 2 SO 4 98 g mol 1 ) are respectively

14 Solutions and Colligative Properties (a) M, 3.77 m (b) 0.03 M, 0.3 m (c) 0.33 M, 7.3 m (d) 33.1 M, 0.7 m 35. Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. If its density is 1.1 g cm 3, the molarity and mole fraction of HCl solution are respectively (a) 1.14 M and (b) M and (c) M and (d) M and A sample of drinking water was found to be severely contaminated with chloroform which is supposed to be carcinogen. If the level of contamination was 15 ppm (by mass) then the molality of chloroform in the water sample is (a) m (b) m (c) m (d) m g of glucose is dissolved in 400 g of water. What is the percentage by mass of glucose solution? (a) 78.7% (b) 7.87% (c) 87.7% (d) 77% 38. A solution is prepared by dissolving certain amount of solute in 500 g of water. If the percentage by mass of a solute in solution is 2.38 then the mass of solute will be (a) g (b) g (c) g (d) 1219 g cm 3 of methyl alcohol is dissolved in 25.2 g of water. The % by mass of methyl alcohol and mole fraction of methyl alcohol are respectively (Given density of methyl alcohol g cm 3 and C 12, H 1, O 16) (a) 1.26% and (b) 12.68% and (c) 12% and (d) 1.26% and cm 3 of benzene is dissolved in 16.8 cm 3 of xylene. What is the % by volume of benzene? (a) 24.43% (b) 4.324% (c) 43.24% (d) 0.43% 41. What is the mole fraction of HCl in aqueous solution of HCl containing 24.8% of HCl by mass? (H 1, Cl 35.5) (a) (b) (c) (d) What is the mole fraction of solute in its 2 molal aqueous solution? (a) (b) 3.47 (c) (d) The molality and molarity of HNO 3 in a solution containing 12.2% HNO 3 are respectively (Given density of HNO g cm 3, H 1, N 14, O 16) (a) 2.01 m and M (b) 1.2 m and 2.6 M (c) m and 2.01 M (d) 20.6 m and 20 M Sulphuric acid is 95.8% by mass. What is the mole fraction and molarity of H 2 SO 4 solution having density 1.91 g cm 3? (H 1, S 32, O 16) (a) and M (b) and M (c) 0.07 and 7.86 M (d) and M 45. Aqueous solution of NaOH is marked 10% (w/w). The density of the solution is g cm 3. The molarity and molality of solution are respectively (Na 23, H 1, O 16) (a) M and 2.77 m (b) 0.27 M and 0.26 m (c) 0.26 M and 0.02 m (d) 20.7 M and 20.6 m 46. Battery acid is 4.22 M aqueous H 2 SO 4 solution, and has density of 1.21 g cm 3. What is the molality of H 2 SO 4? (H 1, S 32, O 16) (a) mol/g (b) mol/kg (c) mol/kg (d) mol/kg 47. A molal solution is one that contains one mole of solute in (a) one litre of the solvent (b) 1000 g of the solvent (c) one litre of the solution (d) 22.4 litres of solution. 48. Which of the following is independent of temperature? (a) Molarity (b) Molality (c) Both (a) and (b) (d) None of these g of NaOH (molar mass 40 g mol 1 ) is dissolved in 500 cm 3 of water. Molality of resulting solution is (a) 0.1 m (b) 0.5 m (c) 1.5 m (d) 1.0 m 50. Molarity of solution depends on (a) temperature (b) nature of solute dissolved (c) mass of solvent (d) pressure. 51. Density of water is 1 g/ml. The concentration of water in mol/litre is (a) 1000 (b) 18 (c) (d) The solutions A and B are 0.1 M and 0.2 M in a substance. If 100 ml of A is mixed with 25 ml of B and there is no change in volume, the final molarity of the solution is (a) 0.15 M (b) 0.18 M (c) 0.30 M (d) 0.12 M

15 58 Objective MHT-CET Chemistry 53. The molarity of a solution that contains 49 g H 3 PO 4 in 2.0 L of a solution is (a) 0.25 M (b) 0.50 M (c) 0.75 M (d) 1.0 M 54. The molarity of the solution containing 7.1 g of Na 2 SO 4 in 100 ml of aqueous solution is (a) 1 M (b) 2 M (c) 0.05 M (d) 0.5 M 55. If 60 cm 3 of ethyl alcohol is dissolved in 300 cm 3 of water then the percentage by volume of ethyl alcohol is (a) 16.66% (b) 15.76% (c) 17.86% (d) 18.96% 56. What volume of 0.8 M solution contains 0.1 mol of the solute? (a) 62.5 ml (b) 100 ml (c) 500 ml (d) 125 ml ml of 3.0 M HNO 3 are mixed with 75 ml of 4.0 M HNO 3. If the volumes are additive, the molarity of the final mixture would be (a) 3.25 M (b) 4.0 M (c) 3.75 M (d) 3.50 M 58. The amount of anhydrous Na 2 CO 3 present in 250 ml of 0.25 M solution is (a) g (b) g (c) 6.0 g (d) g ml of HCl solution requires ml of 0.01 M NaOH solution for complete neutralization. The molarity of HCl solution is (a) M (b) M (c) 0.99 M (d) 9.9 M 60. If 5.85 g of NaCl (molecular weight 58.5 g mol 1 ) is dissolved in water and the solution is made up to 0.5 litre, the molarity of the solution will be (a) 0.2 M (b) 0.4 M (c) 1.0 M (d) 0.1 M 61. If 1 M and 2.5 litre NaOH solution is mixed with another 0.5 M and 3 litre NaOH solution, then molarity of the resultant solution will be (a) 1.0 M (b) 0.73 M (c) 0.80 M (d) 0.50 M 62. To prepare a solution of concentration of 0.03 g/ml of AgNO 3, what amount of AgNO 3 should be added in 60 ml of solution? (a) 1.8 g (b) 0.8 g (c) 0.18 g (d) g 63. A solution of CaCl 2 is 0.5 mol/litre, then the moles of chloride ions in 500 ml will be (a) 0.25 (b) 0.50 (c) 0.75 (d) The sum of mole fractions of A, B and C in an aqueous solution containing 0.2 moles of each A, B and C is (a) 0.6 (b) 0.2 (c) 1.0 (d) Solubility of Gases in Liquids 65. According to Henry's law (a) S KP (b) S K/P (c) S P/K (d) S KP 66. If Henry's law constant for oxygen is mol dm 3 atm 1 and its partial pressure is 0.20 atm, the concentration of dissolved oxygen at NTP will be (a) mol dm 3 (b) mol dm 3 (c) mol dm 3 (d) mol dm The unit of Henry's constant is (a) mol dm 3 atm 1 (b) mol dm 3 atm (c) mol 1 dm 3 atm (d) mol 1 dm 3 atm What is the concentration of dissolved oxygen at 25 C at 1 atmospheric pressure if partial pressure of oxygen is 0.22 atm? The Henry s law constant for oxygen is mol dm 3 atm 1. (a) mol dm 3 (b) mol dm 3 (c) mol dm 3 (d) mol dm What is the Henry s law constant of dissolved O 2 at 10 C at 1 atmospheric pressure, if partial pressure of oxygen is 0.24 atm? The concentration of dissolved oxygen is mol dm 3. (a) mol dm 3 atm 1 (b) mol dm 3 atm 1 (c) mol dm 3 atm 1 (d) data insufficient. 70. Solubility of a gas in a liquid increases with (a) increase of pressure and increase of temperature (b) decrease of pressure and increase of temperature (c) increase of pressure and decrease of temperature (d) decrease of pressure and decrease of temperature. 71. Which law states that the amount of gas dissolved in a given mass of solvent at any temperature is directly proportional to pressure of the gas above the solution? (a) Raoult s law (b) Boyle s law (c) Charles law (d) Henry s law

16 Solutions and Colligative Properties 2.5 Solid Solutions 72. Lead is hardened by the addition of (a) 10-20% aluminium (b) 5-20% manganese (c) 10-20% antimony (d) 20-30% iron 73. Babbitt metal is an alloy of (a) bismuth with tin and copper (b) antimony with tin and copper (c) lead with arsenic (d) aluminium with copper and manganese. 74. Amalgam is (a) a solution of gas in gas (b) a solution of liquid in gas (c) a solution of metals in liquid metal (d) a solution of solid in gas. 75. Stainless steel contains (a) chromium (b) nickel (c) both (a) and (b) (d) none of these. 76. An alloys containing zero temperature coefficient of electrical resistance is (a) manganin (b) duralumin (c) bronze (d) spiegeleisen. 77. An alloy is (a) a solution of solid in liquid (b) a solution of solid in gas (c) a solution of solid in plasma (d) a solution of solid in solid to 20% manganese in iron is (a) ferromanganeous (b) manganin (c) duralumin (d) spiegeleisen. 79. An alloy manganin contains (a) 84% Cu, 12% Mn and 4% Ni (b) 70-80% Mn, 30-20% Fe (c) 10-20% Sb, 90-80% Pb (d) Al, Cu, Mg, and Mn. 2.6 Colligative Properties 80. Which of the following is not a colligative property? (a) Osmotic pressure (b) Elevation in boiling point (c) Vapour pressure (d) Depression in freezing point 81. Which of the following is a colligative property? (a) Osmotic pressure (b) Boiling point (c) Vapour pressure (d) Freezing point 82. Colligative properties are applicable to (a) ideal dilute solutions (b) non-ideal concentrated solutions (c) non-ideal solutions (d) ideal concentrated solutions. 83. The colligative properties of a solutions depend on (a) nature of solute particles present in it (b) nature of solvent used (c) number of solute particles present in it (d) number of moles of solvent only. 84. Which of the following is not the colligative property? (a) DT f (b) DT b (c) K b (d) Osmotic pressure Colligative properties are used for the determination of (a) molar mass (b) equivalent weight (c) arrangement of molecules (d) melting point and boiling point. 86. Which of the following is a colligative property? (a) Conductance of a solution (b) Surface tension of a solution (c) Osmotic pressure of a solution (d) Radioactivity of a solution 87. Which is not a colligative property? (a) Refractive index (b) Lowering of vapour pressure (c) Depression of freezing point (d) Elevation of boiling point 2.7 Lowering of Vapour Pressure 88. Which of the following is incorrect? (a) Relative lowering of vapour pressure is independent of the nature of the solute and the solvent (b) Relative lowering in vapour pressure is a colligative property (c) Vapour pressure of a solution is lower than the vapour pressure of the solvent (d) Relative lowering of vapour pressure is directly proportional to the original pressure. 89. Which one of the following is not an expression for Raoult s law for a solution containing two volatile components A and B? p A and p B are the partial vapour pressures of A and B, p A and p B are the vapour pressures of pure A and B, x A and x B are mole fractions of A and B in solution.

17 60 Objective MHT-CET Chemistry (a) p A p A x A (b) P total p A x A + p B x B (c) Dp p B x B (d) Dp p A x A 90. Lowering of vapour pressure is a colligative property because, it depends on the concentration of (a) volatile solute (b) non-volatile solute (c) volatile solvent (d) non-volatile solvent. 91. At 25 C, the total pressure of an ideal solution obtained by mixing 3 moles of A and 2 moles of B, is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature? (vapour pressure of pure A, at 25 C, is 200 torr.) (a) 180 (b) 160 (c) 16 (d) The relative lowering of vapour pressure is equal to p 1 (a) (b) p p1 Dp p (c) Dp p1 p (d) p 1 Dp 93. The aqueous solution that has the highest value of relative lowering of vapour pressure at a given temperature is (a) 0.1 molal sodium phosphate (b) 0.1 molal barium chloride (c) 0.1 molal sodium chloride (d) 0.1 molal glucose. 94. For dilute solutions, Raoult s law states that (a) lowering of vapour pressure is equal to the mole fraction of the solute (b) relative lowering of vapour pressure is equal to the mole fraction of the solvent (c) relative lowering of vapour pressure of the solvent is equal to the mole fraction of the solute (d) vapour pressure of the solution is equal to the vapour pressure of the solvent. 95. Vapour pressure of dilute aqueous solution of glucose is 750 mm of mercury at 373 K. The mole fraction of solute is (a) 1/76 (b) 1/7.6 (c) 1/38 (d) 1/ Vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985 N/m 2. The vapour pressure of pure water is 3000 N/m 2, the molecular weight of the solute is (a) 90 g mol 1 (b) 200 g mol 1 (c) 180 g mol 1 (d) 380 g mol The vapour pressure of a solution containing kg of solute in 0.1 kg of water at 298 K is mm Hg. Given that the vapour pressure of water at 298 K is mm Hg. The molar mass of the solute will be (a) 69 g mol 1 (b) 95 g mol 1 (c) 60 g mol 1 (d) 90 g mol A solution is prepared from kg of an unknown substance and kg acetone at 313 K. The vapour pressure of pure acetone at this temperature is atm. The vapour pressure of solution is kg mol 1 if the molar mass of substance is (a) atm (b) atm (c) atm (d) atm 99. The vapour pressure of 2.1% solution of a nonelectrolyte in water at 100 C is 755 mm Hg. What is the molar mass of the solute? (a) kg/mol (b) g/mol (c) 5.86 kg/mol (d) g/mol 100. The vapour pressure of water at 20 C is 17 mm Hg. What is the vapour pressure of a solution containing 2.8 g of urea (NH 2 CONH 2 ) in 50 g of water? (N 14, C 12, H 1) (a) mm of Hg (b) mm of Hg (c) mm of Hg (d) mm of Hg 101. In an experiment, g of mannitol was dissolved in 100 g of water. The vapour pressure of water was lowered by mm Hg from mm Hg. The molar mass of mannitol is (a) g/mol (b) kg/mol (c) g/mol (d) g/mol 102. According to the Raoult s law, the relative lowering of vapour pressure is equal to the (a) mole fraction of solvent (b) mole fraction of solute (c) independent of mole fraction of solute (d) molality of solution Partial pressure of solvent in solution of non-volatile solute is given by equation, (a) p x 2 p 1 (b) p 1 xp (c) p x 1 p 1 (d) p 1 x 1 p 104. When partial pressure of solvent in solution of nonvolatile solute is plotted against its mole fraction, nature of graph is (a) a straight line passing through origin (b) a straight line parallel to mole fraction of solvent

18 Solutions and Colligative Properties (c) (d) a straight line parallel to vapour pressure of solvent a straight line intersecting vapour pressure axis Relative lowering of vapour pressure of solution is a (a) property of solute (b) property of solute as well as solvent (c) property of solvent (d) colligative property Vapour pressure of solution of a non-volatile solute is always (a) equal to the vapour pressure of pure solvent (b) higher than vapour pressure of pure solvent (c) lower than vapour pressure of pure solvent (d) constant The vapour pressure at equilibrium of a liquid in a closed vessel depends on (a) pressure (b) concentration (c) temperature (d) volume The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, its vapour pressure at 300 K will be (a) 0.8 atm (b) 0.2 atm (c) 0.4 atm (d) 0.6 atm 109. At 300 K when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be (a) (b) (c) (d) If p 1 and p are the vapour pressures of a solvent and its solution respectively, and if x 1 and x 2 are the mole fractions of the solvent and solute respectively then (a) p p 1 x 1 (b) p p 1 x 2 (c) p 1 p x 2 (d) p p 1 (x 1 /x 2 ) 111. The vapour pressure of pure liquid A is 0.80 atm. On mixing a non-volatile solute B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is (a) (b) 0.25 (c) 0.50 (d) The vapour pressure of two liquids P and Q are 80 and 60 torr respectively. The total vapour pressure of solution obtained by mixing 3 moles of P and 2 moles of Q would be (a) 140 torr (b) 20 torr (c) 68 torr (d) 72 torr g of urea (mol. wt. 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is p 1, the vapour pressure of solution is (a) 0.10 p 1 (b) 1.10 p 1 (c) 0.90 p 1 (d) 0.99 p The vapour pressure of a solvent is decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is 20 mm of mercury? (a) 0.8 (b) 0.6 (c) 0.4 (d) The vapour pressure of a pure liquid A is 70 torr at 27 C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and the total vapour pressure of the solution is 84 torr at 27 C. The vapour pressure of pure liquid B at 27 C is (a) 14 torr (b) 56 torr (c) 70 torr (d) 140 torr 116. At 40 C the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be (a) 140 mm Hg (b) 110 mm Hg (c) 220 mm Hg (d) 100 mm Hg 117. One mole of sugar is dissolved in two moles of water. The vapour pressure of the solution relative to that of pure water is (a) 2/3 (b) 1/3 (c) 3/2 (d) 1/ The vapour pressure of benzene at 80 C is lowered by 10 mm by dissolving 2 g of a non-volatile substance in 78 g of benzene. The vapour pressure of pure benzene at 80 C is 750 mm. The molecular mass of the substance will be (a) 15 g mol 1 (b) 150 g mol 1 (c) 1500 g mol 1 (d) 148 g mol Vapour pressure of CCl 4 at 25 C is 143 mm of Hg. 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in 100 ml CCl 4. Find the vapour pressure of the solution. (Density of CCl g/cm 3 ) (a) mm (b) mm (c) mm (d) mm 120. The vapour pressure of a solution increases when (a) the temperature is raised (b) the volume is increased (c) the number of moles of the solute is increased (d) the temperature is lowered.

19 62 Objective MHT-CET Chemistry 121. Pressure cooker reduces cooking time for food because (a) heat is more evenly distributed in the cooking space (b) boiling point of water involved in cooking is increased (c) the higher pressure inside the cooker crushed the food material (d) cooking involves chemical changes helped by a rise in temperature The relative lowering of the vapour pressure is equal to the ratio between the number of (a) solute molecules and solvent molecules (b) solute molecules and the total molecules in the solution (c) solvent molecules and the total molecules in the solution (d) solvent molecules and the total number of ions of the solute The vapour pressure lowering caused by the addition of 100 g of sucrose (molecular mass 342) to 1000 g of water if the vapour pressure of pure water at 25 C is 23.8 mm Hg is (a) 1.25 mm Hg (b) mm Hg (c) 1.15 mm Hg (d) mm Hg 2.8 Boiling Point Elevation 124. Boiling point of water is K. What is K b of water, if 0.15 molal aqueous solution of a substance boils at K? (a) 0.09 K kg mol 1 (b) 0.6 K kg mol 1 (c) 6.0 K kg mol 1 (d) 60 K kg mol K b is the elevation in boiling point produced when 1 mol of solute is dissolved in (a) one litre solvent (b) 1000 g solvent (c) 500 ml solvent (d) 0.5 kg solvent If 0.15 molal solution of a substance boils at K the molal elevation constant of water is (Given boiling point of water K) (a) 0.35 K kg mol 1 (b) 5.3 K kg mol 1 (c) 0.53 K kg mol 1 (d) K kg mol A solution of a substance on dissolving in benzene boils at K. If K b for benzene is 2.53 K kg mol 1 and boiling point of pure benzene is K, then the molality of the solution is (a) m (b) m (c) 0.04 m (d) m 128. A solution is prepared by dissolving kg ammonia in 400 g of water. The elevation in boiling point if K b for water is 0.52 K kg mol 1 (N 14, H 1, O 16) is (a) 145 C (b) 1.45 K (c) 14.5 K (d) 45.1 K 129. A solution containing 1.21 g of camphor (molar mass 152 g mol 1 ) in g of acetone boils at K. If the boiling point of pure acetone is K, the molal elevation constant for acetone will be (a) 167 K kg mol 1 (b) 1.67 K mol 1 (c) K kg mol 1 (d) K kg mol g of sulphur is dissolved in 100 g of CS 2. This solution boils at K. What is the molecular formula of sulphur in solution? The boiling point of CS 2 is K. (Given : K b for CS K kg mol 1 and atomic mass of S 32.) (a) S 2 (b) S 4 (c) S 8 (d) None of these A solution prepared by dissolving certain amount of compound in 31.8 g of CCl 4 has a boiling point of K higher than that of pure CCl 4. If the molar mass of compound is 128 g mol 1, the mass of the solute dissolved will be (Given K b for CCl K kg mol 1 ) (a) 317 g (b) kg (c) g (d) 31.7 g 132. Boiling point of water at 750 mm Hg is C. How much sucrose is to be added to 500 g of water such that it boils at 100 C? Molal elevation constant for water is 0.52 K kg mol 1. (a) kg (b) g (c) kg (d) g 133. A solution containing g of naphthalene (molar mass g mol 1 ) in g of CCl 4 gives a boiling point elevation of K. While a solution of g of unknown solute in the same mass of the solvent gives a boiling point elevation of k. The molar mass of the unknown solute is (K b for CCl K kg mol 1 ) (a) kg/mol (b) kg/mol (c) kg/mol (d) kg/mol 134. The mass in grams of an impurity of molar mass 100 g mol 1 which would be required to raise the boiling point of 50 g of chloroform by 0.30 K (K b for chloroform 3.63 K kg mol 1 ) is (a) g (b) g (c) 43.1 g (d) 42.3 g

20 Solutions and Colligative Properties 135. Boiling point of a solvent is 80.2 C. When g of the solute of molar mass g mol 1 was dissolved in 75 g of the solvent, the boiling point of the solution was found to be C. What is the molal elevation constant? (a) 25.4 K kg/mol (b) 2.54 K g/mol (c) 25.4 K g/mol (d) 2.54 K kg/mol 136. When NaCl is added to water (a) freezing point is raised (b) boiling point is depressed (c) freezing point does not change (d) boiling point is raised Molal elevation constant is elevation in boiling point produced by (a) 1 g of solute in 100 g of solvent (b) 100 g of solute in 1000 g of solvent (c) 1 mole of solute in one litre of solvent (d) 1 mole of solute in one kg of solvent The determination of molar mass from elevation in boiling point is called as (a) cryoscopy (b) osmometry (c) ebullioscopy (d) spectroscopy If mass is expressed in gram then K b is given by M T W (a) 2D b W1 (b) (c) (d) W D Tb W1 M2 M2D Tb W W2 W D Tb W2 M Unit of boiling point elevation constant (K b ) is (a) kg mol 1 (b) K mol 1 (c) g mol 1 (d) K kg mol The rise in the boiling point of a solution containing 1.8 g of glucose in 100 g of a solvent is 0.1 C. The molal elevation constant of the liquid is (a) 0.01 K/m (b) 0.1 K/m (c) 1 K/m (d) 10 K/m 142. The molal elevation of boiling point constant for water is C kg mol 1. When 0.1 mole of sugar is dissolved in 200 ml of water, the solution boils under a pressure of one atmosphere at (a) C (b) C (c) C (d) C An aqueous solution containing 1 g of urea boils at C. The aqueous solution containing 3 g of glucose in the same volume will boil at (molecular weights of urea and glucose are 60 and 180 respectively) (a) C (b) C (c) C (d) 100 C 144. If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by C than that of the pure solvent, the molecular weight of the substance (molal elevation of boiling point constant for the solvent is 2.16 C) is (a) 1.01 g mol 1 (b) 10 g mol 1 (c) 1000 g mol 1 (d) 100 g mol A liquid is in equilibrium with its vapours at its boiling point. On average, the molecules in the two phases have equal (a) intermolecular forces (b) potential energy (c) total energy (d) kinetic energy Which of the following aqueous solutions containing 10 g of solute in each case has the highest boiling point? (a) NaCl solution (b) KCl solution (c) Sugar solution (d) Glucose solution 147. Elevation in boiling point was 0.52 C when 6 g of a compound X was dissolved in 100 g of water. Molecular weight of X is (K b for water is 0.52 K kg mol 1 ). (a) 120 g mol 1 (b) 60 g mol 1 (c) 180 g mol 1 (d) 600 g mol At higher altitudes the boiling point of water is lower because (a) atmospheric pressure is low (b) temperature is low (c) atmospheric pressure is high (d) temperature is high The boiling point of C 6 H 6, CH 3 OH, C 6 H 5 NH 2 and C 6 H 5 NO 2 are 80 C, 65 C, 184 C and 212 C respectively. Which will show highest vapour pressure at room temperature? (a) C 6 H 6 (b) CH 3 OH (c) C 6 H 5 NH 2 (d) C 6 H 5 NO The boiling point of 0.1 molal aqueous solution of urea is C at 1 atm. The molal elevation constant of water (in C kg mol 1 ) is (a) 1.8 (b) 0.18 (c) 18 (d) 18.6

21 64 Objective MHT-CET Chemistry 151. If the solution boils at a temperature T 1 and the solvent at a temperature T 2, the elevation of boiling point is given by (a) T 1 + T 2 (b) T 1 T 2 (c) T 2 T 1 (d) T 1 T The molal elevation constant is the ratio of the elevation in boiling point to (a) molarity (b) molality (c) mole fraction of solute (d) mole fraction of solvent Elevation of boiling point was 0.52 C, when 6 g of compound X was dissolved in 100 g of H 2 O. Molar mass of X is (K b of water is 5.2 C per 100 g water) (a) 120 g mol 1 (b) 60 g mol 1 (c) 600 g mol 1 (d) 180 g mol The elevation of boiling point method is used for the determination of molecular weight of (a) non-volatile and soluble solute (b) non-volatile and insoluble solute (c) volatile and soluble solute (d) volatile and insoluble solute When a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in (a) an increase in the boiling point of the solution (b) a decrease in the boiling point of the solvent (c) the solution having a higher freezing point than the solvent (d) the solution having a lower osmotic pressure than the solvent When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by 1 C, then molecular mass of the solute is (K b for benzene 2.53 K kg mol 1 ) (a) 223 g (b) 233 g (c) 243 g (d) 253 g 157. The value of K b for a solvent is X K kg mol 1. A 0.2 m solution of a non-electrolyte in this solvent will boil at (Given : boiling point of solvent A C). (a) (A + X) C (b) A + X 10 C (c) A + X 5 C (d) A + X 5 K 158. At certain hill-station pure water boils at C. If K b for water is C kg mol 1, the boiling point of 0.69 m solution of urea will be (a) C (b) 103 C (c) C (d) Unpredictable 2.9 Freezing Point Depression 159. Which of the following solutions shows maximum depression in freezing point? (a) 0.5 M Li 2 SO 4 (b) 1 M NaCl (c) 0.5 M Al 2 (SO 4 ) 3 (d) 0.5 M BaCl The unit of freezing point depression constant is (a) K mol 1 (b) K kg 1 mol 1 (c) K kg mol 1 (d) K kg The freezing point of equimolal aqueous solution will be highest for (a) + C 6 H 5 NH 3 Cl (b) La(NO 3 ) 3 (c) glucose (d) Ca(NO 3 ) A 0.5 molal solution of ethylene glycol in water is used as coolant in a car. If K f for water is 1.86 K kg mol 1 the mixture will freeze at (a) 0.93 C (b) 0.93 C (c) 1.86 C (d) 1.86 C g of a substance dissolved in 50 g of water lowered the freezing point by 1.2 K. If the molal depression constant for water is 1.86 K kg mol 1, the molar mass of the substance, will be (a) g mol 1 (b) g mol 1 (c) g mol 1 (d) 15.7 kg mol kg of a substance (molar mass kg mol 1 ) dissolved in kg of benzene lowered the freezing point of benzene by K. The molal depression constant for benzene is (a) 512 K kg mol 1 (b) 5.12 K kg mol 1 (c) 51.2 K kg mol 1 (d) none of these The observed depression in the freezing point of water for a particular solution is K. The molality of the solution if molal depression constant for water is 1.86 K kg mol 1 will be (a) 0.4 m (b) 4.67 m (c) m (d) 4 m g of urea when dissolved in 98.5 g of certain solvent decreases its freezing point by K g of unknown compound when dissolved in 86.0 g of the same solvent depresses the freezing point by 0.34 K. The molar mass of the unknown compound is (Urea NH 2 CONH 2, N 14, C 12, O 16, H 1) (a) 60 g mol 1 (b) 68 g mol 1 (c) 78 g mol 1 (d) 40 g mol 1

22 Solutions and Colligative Properties 167. A solution of glucose (C 6 H 12 O 6 ) was prepared by dissolving certain amount of glucose in 100 g of water. The depression in freezing point was K. If molal depression constant for water is 1.86 K kg mol 1, the mass of glucose dissolved will be (C 12, H 1, O 16) (a) g (b) kg (c) 39.6 g (d) 39.6 kg 168. An aqueous solution containing kg of non-volatile compound in 0.1 kg of water freezes at K. The molar mass of the compound is (K f for water 1.86 K kg mol 1. (a) g mol 1 (b) g mol 1 (c) g mol 1 (d) g mol The freezing point of solution prepared by dissolving 4.5 g of glucose (Molar mass 180 g mol 1 ) in 250 g of bromoform is (Given : Freezing point of bromoform 7.8 C and K f for bromoform 14.4 K kg mol 1 ) (a) 6.36 C (b) 4.23 C (c) 3.66 C (d) 0.06 C 170. A temperature at which the vapour pressure of a solid is equal to the vapour pressure of liquid is called (a) elevation of boiling point (b) freezing point (c) melting point (d) depression of freezing point 171. The freezing point of 1 percent solution of lead nitrate in water will be (a) below 0 C (b) 0 C (c) 1 C (d) 2 C 172. How much polystyrene of molecular weight 9000 would have to be dissolved in 100 grams of C 6 H 6 to lower its freezing point by 1.05 C? (K f C 6 H 6 4.9) (a) 19.3 g (b) 193 g (c) 38.6 g (d) 77.2 g 173. Relationship between K f, m and DT f can be written as (a) DT f K f /m (b) DT f K f m (c) DT f K f (d) DT f m 174. Which will show maximum depression in freezing point when concentration is 0.1 M? (a) NaCl (b) Urea (c) Glucose (d) K 2 SO In cold countries, ethylene glycol is added to water in the radiators of cars during winter. It results in (a) (b) (c) (d) reducing viscosity reducing specific heat reducing freezing point reducing boiling point The molar mass of the solute using depression of freezing point may be calculated using 10K f W2 100K f W2 (a) M2 (b) M D 2 Tf m D TW f 1 K f W1 1000K f W2 (c) M2 (d) M D 2 TW f 2 D TW f The freezing point of 0.05 molal solution of nonelectrolyte in water is (K b 1.86 C kg mol 1 ) (a) 1.86 C (b) 0.93 C (c) C (d) 0.93 C 178. What is the molality of solution of a certain solute in a solvent, if there is a freezing point depression of C, and the freezing point constant is 18.4? (a) 0.01 m (b) 1 m (c) m (d) 100 m 179. If all the following four compounds were sold at the same price, which would be cheapest for preparing an antifreeze solution for a car radiator? (a) Methanol (b) Ethanol (c) Ethylene glycol (d) Glycerol 180. Equimolal solutions of A and B show depression of freezing point in the ratio of 2 : 1. A remains in normal state. The state of B in the solution is (a) normal (b) dissociated (c) associated (d) unpredictable A solution containing 6.8 g of a non-ionic solute in 100 g water was found to freeze at 0.93 C. The freezing point depression constant of water is The molar mass of the solute is (a) 13.6 g mol 1 (b) 34 g mol 1 (c) 68 g mol 1 (d) 136 g mol Which of the following aqueous solutions has minimum freezing point? (a) 0.01 m NaCl (b) m C 2 H 5 OH (c) m MgI 2 (d) m MgSO When mercuric iodide is added to the aqueous solution of potassium iodide, the (a) freezing point is lowered (b) freezing point is raised (c) freezing point does not change (d) boiling point does not change.

23 66 Objective MHT-CET Chemistry 184. An aqueous solution freezes at C. The boiling point of the same solution is (K f 1.86 K m 1, K b K m 1 ) (a) C (b) C (c) (d) C 185. An aqueous solution of a non-electrolyte boils at C. The freezing point of the solution will be (K f 1.86 K m 1, K b K m 1 ) (a) 0 C (b) 1.86 C (c) 1.86 C (d) 0.52 C 186. A solution of urea (mol. mass 56 g mol 1 ) boils at C at the atmospheric pressure. If K f and K b for water are 1.86 and K kg mol 1 respectively the above solution will freeze at (a) 6.54 C (b) 6.54 C (c) C (d) C 187. After adding a solute freezing point of solution decreases to Calculate DT b if K f 1.86 and K b (a) (b) (c) 1.86 (d) Which of the following will have the highest freezing point at one atmosphere? (a) 0.1 M NaCl solution (b) 0.1 M sugar solution (c) 0.1 M BaCl 2 solution (d) 0.1 M FeCl 3 solution g of substance x dissolved in 100 g of water freezes at 0.93 C. The molar mass of x is (K f 1.86 K m 1 ) (a) 60 g mol 1 (b) 120 g mol 1 (c) 180 g mol 1 (d) 140 g mol What is the molality of ethyl alcohol (mol. wt. 46) in aqueous solution which freezes at 10 C? (K f for water 1.86 K molality 1 ) (a) m (b) m (c) m (d) m 191. If K f value of H 2 O is 1.86, the value of DT f for 0.1 m solution of non-volatile solute is (a) 18.6 (b) (c) 1.86 (d) % solution of Ca(NO 3 ) 2 has the freezing point (a) 0 C (b) less than 0 C (c) greater than 0 C (d) 273 K 193. If a substance exists as trimer in the solution, then which of the following alternative is possible for depression in freezing point of m molal solution? mk f mk (a) (b) f 3 4 (c) mk f 5 (d) mk f Osmosis and Osmotic Pressure g of a solute is dissolved in 0.1 L of a solvent which develops an osmotic pressure of 1.23 atm at 27 C. The molecular mass of the substance is (a) g mol 1 (b) 120 g mol 1 (c) 430 g mol 1 (d) 102 g mol Solutions having same osmotic pressure are known as (a) hypotonic (b) hypertonic (c) isotonic (d) isomeric The osmotic pressure of a dilute solution is directly proportional to the (a) diffusion rate of the solute (b) concentration (c) boiling point (d) flow of solvent from a concentrated to a dilute solution At a given temperature isotonic solutions have the same (a) density (b) volume (c) normality (d) molar concentration The osmotic pressure is expressed in units of (a) MeV (b) calories (c) cm (d) atmosphere The scientist, who discovered the phenomenon of osmosis in natural membranes is (a) Raoult F. Marie (b) E. Otto (c) van't Hoff (d) Abbe Nollet What happens when blood cells are placed in pure water? (a) The fluid in blood cells rapidly moves into water (b) The water molecules rapidly move into blood cells (c) The blood cells dissolve in water (d) No change takes place Calculate the osmotic pressure of a solution containing g glucose in 0.1 litre of the solution at 298 K? (a) atm (b) atm (c) atm (d) atm 202. The osmotic pressure of 4.5 g of glucose (molar mass 180 g mol 1 ) dissolved in 100 ml of water at 298 K (R L atm mol 1 K 1 ) is (a) atm (b) 61.6 atm (c) atm (d) atm

24 Solutions and Colligative Properties 203. A solution containing 17.8 g L 1 of cane sugar (molar mass 342 g mol 1 ) has an osmotic pressure 1.2 atm. What is the temperature of the solution? (R L atm mol 1 K 1 ). (a) C (b) K (c) K (d) C g of glucose dissolved in one litre of water has an osmotic pressure 4.91 atm at 303 K. If the osmotic pressure of the glucose solution is 1.5 atm at the same temperature, what would be its concentration? (Molar mass of glucose is 180 g mol 1 ) (a) 0.5 M (b) M (c) 5 M (d) None of these 205. A solution has an osmotic pressure of Nm 2 at 300 K. The volume containing 1 mole of solute if solution of same solute has an osmotic pressure of Nm 2 and contains 1 mole of solute in 10.5 m 3 is (a) 7.59 m 3 (b) 7.59 L (c) 7.59 cm 3 (d) 75.9 m What is the mass of sucrose in its 1 L solution (molar mass 342 g mol 1 ) which is isotonic with kg L 1 of urea (NH 2 CONH 2 )? (Given : atomic masses H 1, C 12, N 14, O 16 in g mol 1 ) (a) kg (b) g (c) g (d) g 207. Osmotic pressure of a solution containing kg of protein per m 3 of solution is Pa at 37 C. What is the molar mass of protein? (R J K 1 mol 1 ) (a) kg mol 1 (b) kg mol 1 (c) kg mol 1 (d) kg mol At 298 K, 1000 cm 3 of a solution containing 4.34 g of solute shows osmotic pressure of 2.55 atm. What is the molar mass of solute? (R L atm K 1 mol 1 ) (a) g mol 1 (b) g mol 1 (c) g mol 1 (d) g mol What is the volume of a solution containing 34.2 g of cane sugar (molar mass 342 g mol 1 ) which has an osmotic pressure 2.42 atm at 20 C? (a) L (b) L (c) L (d) L 210. A solution of particular amount of organic substance of molar mass 196 g mol 1 dissolved in 2 litres of water gave an osmotic pressure of 0.54 atm at 12 C. The mass of the solute dissolved is [R L atm K 1 mol 1 ] (a) 10.5 g (b) 9.04 kg (c) 9.04 g (d) 0.9 kg Which of the following 0.1 M aqueous solutions will exert highest osmotic pressure? (a) NaCl (b) BaCl 2 (c) MgSO 4 (d) Al 2 (SO 4 ) In osmosis (a) solvent molecules pass from high concentration of solute to low concentration. (b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute. (c) solute molecules pass from low concentration to high concentration (d) solute molecules pass from high concentration to low concentration A solution contains non-volatile solute of molecular mass M B. Which of the following can be used to calculate molecular mass of the solute in terms of osmotic pressure? (W B Mass of solute, V Volume of solution and p Osmotic pressure) (a) M B W B VRT (b) M B W B RT p pv (c) M B W B V p RT (d) M B WB p V RT 214. A membrane which allows solvent molecules but not the solute molecules to pass through it is called as (a) semipermeable membrane (b) permeable membrane (c) filter membrane (d) porous membrane Which inorganic precipitate acts as semipermeable membrane? (a) Calcium sulphate (b) Barium oxalate (c) Nickel phosphate (d) Copper ferrocyanide 216. The osmotic pressure of a solution increases if (a) temperature is decreased (b) solution constant is increased (c) number of solute molecule is increased (d) volume is increased The solution in which the blood cells retain their normal form are to the blood. (a) isotonic (b) isomotic (c) hypertonic (d) equinormal 218. In equimolar solutions of glucose, NaCl and BaCl 2, the order of osmotic pressure is (a) Glucose > NaCl > BaCl 2 (b) NaCl > BaCl 2 > Glucose

25 68 Objective MHT-CET Chemistry (c) (d) BaCl 2 > NaCl > Glucose Glucose > BaCl 2 > NaCl 219. A solution having a higher osmotic pressure than another solution, is called a (a) hypotonic solution (b) isotopic solution (c) isotonic solution (d) hypertonic solution Water transportation in plants takes place by the phenomenon of (a) diffusion (b) osmosis (c) reverse osmosis (d) reverse diffusion The solution containing 4.0 g of a polyvinyl chloride polymer in 1 litre of dioxane was found to have an osmotic pressure atmosphere at 300 K, the value of R used is L atm mol 1 K 1. The molecular mass of the polymer was found to be (a) (b) (c) (d) If mole fraction of the solvent in a solution decreases then (a) vapour pressure of solution increases (b) boiling point decreases (c) osmotic pressure increases (d) none of these The concentration in g/l of a solution of cane sugar (molar mass 342 g mol 1 ) which is isotonic with a solution containing 6 g of urea (molar mass 60 g mol 1 ) per litre is (a) 3.42 (b) 34.2 (c) 5.7 (d) The osmotic pressure of 10% solution of cane sugar at 69 C in atmospheres is (a) 724 (b) 824 (c) 8.21 (d) The relationship between osmotic pressure at 273 K when 10 g glucose (P 1 ), 10 g urea (P 2 ) and 10 g sucrose (P 3 ) are dissolved in 250 ml of water is (a) P 1 > P 2 > P 3 (b) P 3 > P 1 > P 2 (c) P 2 > P 1 > P 3 (d) P 2 > P 3 > P Pressure at which reverse osmosis starts is (a) very high pressure (b) atmospheric pressure (c) osmotic pressure (d) vapour pressure Osmotic pressure of a solution containing 3 g of glucose in 60 g of water at 15 C is (density 1 g/ml, molecular weight of glucose 180 g/mol) (a) 0.34 atm (b) 0.65 atm (c) 6.56 atm (d) 5.57 atm 228. A 5% solution of cane sugar (mol. wt. 342) is isotonic with 1% solution of a substance X. The molecular weight of X is (a) 34.2 (b) (c) 68.4 (d) The hard shell of an egg is dissolved in acetic acid, and then egg is placed in saturated solution of NaCl. Choose the correct statement. (a) The egg will shrink. (b) The egg will become harder. (c) The egg will swell. (d) There will be no change in the size of egg Two solutions of KNO 3 and CH 3 COOH are prepared separately. Molarity of both is 0.1 M and osmotic pressures are P 1 and P 2 respectively. The correct relationship between the osmotic pressures is (a) P 2 > P 1 (b) P 2 P 1 (c) P 1 > P 2 (d) P1 P 2 P1 P2 P1+ P A plant cell shrinks when it is kept in (a) a hypotonic solution (b) a hypertonic solution (c) a solution isotonic with the cell sap (d) water g of a solute was dissolved in 500 ml of water and osmotic pressure of the solution was found to be 600 mm of Hg at 15 C. Molecular weight of the solute is (a) 1000 g mol 1 (b) 1200 g mol 1 (c) 1400 g mol 1 (d) 1800 g mol Which statement is incorrect for osmotic pressure (p), volume (V) and temperature (T)? (a) (b) (c) (d) p 1 if T is constant V p V if T is constant p T if V is constant pv is constant if T is constant 234. The osmotic pressure of a decinormal solution of BaCl 2 in water is (a) inversely proportional to its Celsius temperature (b) inversely proportional to its absolute temperature (c) directly proportional to its Celsius temperature (d) directly proportional to its absolute temperature After swimming for a long time in salt water the skin of one s finger tips wrinkles. Which one of the following property is responsible for this observation? (a) Osmosis (b) Dialysis (c) Electrodialysis (d) Coagulation

26 Solutions and Colligative Properties 236. The osmotic pressure of which solution is maximum? (consider that deci-molar solution of each is 90% dissociated) (a) Aluminium sulphate (b) Barium chloride (c) Sodium sulphate (d) A mixture of equal volumes of (b) and (c) 237. The osmotic pressure of 0.4% urea solution is 1.66 atm and that of 3.42% solution of sugar is 2.46 atm. When both the solution are mixed, then the osmotic pressure of the resultant solution will be (a) 1.64 atm (b) 2.46 atm (c) 2.06 atm (d) 0.82 atm 238. If a 0.1 M solution of glucose (mol. wt. 180 g mol 1 ) and 0.1 M solution of urea (mol. wt. 60 g mol 1 ) are placed on the two sides of a semipermeable membrane to equal heights, then it will be correct to say that (a) there will be no net movement across the membrane (b) glucose will flow across the membrane into urea solution (c) urea will flow across the membrane into glucose solution (d) water will flow from urea solution into glucose solution Solutions containing 1.63 g of boric acid in 450 ml and 20 g of sucrose (molecular mass 342) per litre are isotonic. The molar mass of boric acid is (a) (c) (b) (d) The osmotic pressure of a solution at 276 K is 2.5 atm. Its osmotic pressure at 546 K under similar conditions will be (a) 0.5 atm (b) 1.0 atm (c) 2.5 atm (d) 5.0 atm M solution of urea is isotonic with (a) 0.5 M NaCl solution (b) 0.5 M sugar solution (c) 0.5 M BaCl 2 solution (d) 0.5 M solution of benzoic acid in benzene The osmotic pressure of a solution at 0 C is 4 atmospheres. What will be its osmotic pressure at 273 C under similar conditions? (a) 4 atm (b) 2 atm (c) 8 atm (d) 1 atm The molar mass (M 2 ) of W 2 g solute and the osmotic pressure (p) of the solution prepared in V litres by the solute at temperature T has the following relationship (a) M 2 WRT 2 (b) M pv 2 WR 2 pt (c) M 2 mrt (d) M 2 RT p p 244. If osmotic pressure of 1 M of the following in water can be measured, which one will show the maximum osmotic pressure? (a) AgNO 3 (b) MgCl 2 (c) (NH 4 ) 3 PO 4 (d) Na 2 SO Which of the following has the highest osmotic pressure? (a) M/10 HCl (b) M/10 urea (c) M/10 BaCl 2 (d) M/10 glucose 246. According to van t Hoff Avogadro s law, volume occupied by a solution is (a) directly proportional to molar mass of solute (b) inversely proportional to mass of solute (c) directly proportional to number of moles of solute (d) inversely proportional to number of molecules of solute Abnormal Molecular Masses 247. Abnormal colligative properties are observed only when the dissolved non-volatile solute in a given dilute solution (a) is a non-electrolyte (b) offers an intense colour (c) associates or dissociates (d) offers no colour The expression to compute molar mass of a solute from the elevation of boiling point of a solvent is (where the various symbols have their usual meanings) K W T b 1 b W2 (a) M2 (b) M D 2 D Tb W2 Kb W1 (c) M K b W T 2 b W1 2 (d) M2 D D T W K W b Abnormal molar mass is produced by (a) association of solute (b) dissociation of solute (c) both association and dissociation of solute (d) separation by semipermeable membrane. b 2

27 70 Objective MHT-CET Chemistry 250. How many grams of KCl should be added to 1000 g of water, so that the freezing point reduces to 10 C? (K f for water 1.86 C kg mol 1 molar mass of KCl 74.5 g/mole). (a) 74.5 g (b) 745 g (c) 268 g (d) g 251. Acetic acid dissolved in benzene shows a molecular mass of (a) 30 (b) 60 (c) 120 (d) C 6 H 5 COOH (C 6 H 5 COOH) 2 and 2CH 3 COOH (CH 3 COOH) 2 represents (a) association (b) polymerisation (c) condensation (d) evaporation The molecular mass of acetic acid dissolved in water is 60 and when dissolved in benzene it is 120. This difference in behaviour of CH 3 COOH is because (a) water prevents association of acetic acid (b) acetic acid does not dissolve fully in water (c) acetic acid fully dissolves in benzene (d) acetic acid does not ionize in benzene van t Hoff Factor 254. The van t Hoff factor i for a dilute aqueous solution of sucrose is (a) zero (b) 1.0 (c) 1.5 (d) If p obs observed colligative property and p cal theoretical colligative property assuming normal behaviour of solute then van t Hoff factor (i) is given by (a) i p obs p cal (b) i p obs + p cal (c) i p obs p cal (d) i p pobs 256. If a is the degree of dissociation of Na 2 SO 4 the van t Hoff factor (i) used for calculating molecular mass is (a) 1 + a (b) 1 a (c) 1 + 2a (d) 1 2a m aqueous solution of KCl freezes at C. The van t Hoff factor and observed osmotic pressure of solution at 0 C (K f 1.86 K kg mol 1 ) are respectively (a) 1.83, 8.19 atm (b) 8.13, 9.18 atm (c) 3.18, 19.8 atm (d) 3.81, 89.1 atm molal aqueous solution of K 3 [Fe(CN) 6 ] freezes at C. What is the percentage dissociation of solute, K f for water is 1.86 K kg mol 1? cal (a) 87% (b) 78% (c) 89% (d) 98% g of benzoic acid when dissolved in 100 g of benzene lowers its freezing point by 1.62 K. What is the degree of association of benzoic acid if it, forms dimers in benzene, K f for Benzene is 4.9 K kg mol 1 (C 12, H 1, O 16)? (a) 99.16% (b) 9.9% (c) 91.6% (d) 9.16% ml of glacial acetic acid with density 1.06 g ml 1 is dissolved in 1 kg water and the solution froze at C. What is the van t Hoff factor if K f for water is 1.86 K kg mol 1? (a) (b) 2.41 (c) (d) The van t Hoff factor i for a 0.2 molal aqueous solution of urea is (a) 0.2 (b) 0.1 (c) 1.2 (d) van t Hoff factor is (a) less than one in case of dissociation (b) more than one in case of association (c) always less than one (d) less than one in case of association Which of the following compounds has van t Hoff factor i equal to 2 for dilute solution? (a) K 2 SO 4 (b) NaHSO 4 (c) Sugar (d) MgSO van t Hoff factor (i) is the ratio of (a) observed molar mass to theoretical molar mass (b) observed value of colligative property to theoretical value (c) theoretical value of colligative property to observed value (d) none of these The van t Hoff factor will be highest for (a) sodium chloride (b) magnesium chloride (c) sodium phosphate (d) urea The depression in freezing point for 1 M urea, 1 M glucose and 1 M NaCl are in the ratio (a) 1 : 1 : 2 (b) 3 : 2 : 2 (c) 2 : 1 : 1 (d) 1 : 1 : The degree of dissociation (a) of a weak electrolyte A x B y is related to van t Hoff factor (i) by the expression i 1 i 1 (a) a (b) a ( x+ y 1) x+ y+ 1 x (c) a + y 1 i 1 x (d) a + y + 1 i 1

28 Solutions and Colligative Properties 268. The van t Hoff factor i for 0.2 m aqueous glucose solution is (a) 0.2 (b) 0.4 (c) 0.6 (d) The van t Hoff factor calculated from association data is always than calculated from dissociation data. (a) less (b) more (c) same (d) more or less 270. If van t Hoff factor for dissolution of Ca(NO 3 ) 2 in water is 2.5, then degree of dissociation is (a) 25% (b) 50% (c) 75% (d) 84% 271. The substance A when dissolved in solvent B gave molar mass corresponding to A 3. The van t Hoff factor will be 1 (a) (b) (c) 2 (d) Which of the following compounds corresponds to van t Hoff factor (i) to be equal to 2 for dilute solution? (a) Na 2 SO 4 (b) AlCl 3 (c) Glucose (d) BaSO Which one of the following salt, will have the same value of van t Hoff factor (i) as that of K 4 [Fe(CN) 6 ]? (a) Fe 2 (SO 4 ) 3 (b) NaNO 3 (c) Ca(NO 3 ) 2 (d) K 2 SO The van t Hoff factor 0.1 M CaCl 2 solution is The degree of dissociation is (a) 61% (b) 87% (c) 100% (d) 54% 275. What will be the ratio of any of the colligative properties of 1.0 m aqueous solutions of NaCl, Na 2 SO 4 and K 4 [Fe(CN) 6 ] [Assume that solute completely (100%) dissociates in the solution] (a) 2 : 3 : 4 (b) 1 : 2 : 4 (c) 2 : 3 : 5 (d) 1 : 3 : 5 LEVEL During osmosis, flow of water through a semipermeable membrane is (a) from solution having lower concentration only (b) from solution having higher concentration only (c) from both sides of semipermeable membrane with equal flow rates (d) from both sides of semipermeable membrane with unequal flow rates. 2. Find the percentage of aqueous cane sugar solution which will have the same freezing point as that of 3% aqueous solution of urea (molecular weight of urea 60 g mol 1, and molecular weight of cane sugar 342 g mol 1 ). (a) 17.1% (b) 16.6% (c) 77% (d) 20% 3. Which of the following salts has the same value of vant Hoff factor as that of K 3 [Fe(CN) 6 ]? (a) Na 2 SO 4 (b) Al(NO 3 ) 3 (c) Al 2 (SO 4 ) 3 (d) Fe 3 O 4 4. If equimolar solutions of CaCl 2 and AlCl 3 in water have boiling point of T 1 and T 2 respectively then (a) T 1 > T 2 (b) T 2 > T 1 (c) T 1 T 2 (d) can't say 5. If the osmotic pressure of a dilute solution is Pa at 273 K then its osmotic pressure at 283 K will be (a) Nm 2 (b) Nm 2 (c) Nm 2 (d) Nm 2 6. The solubility of a gas in liquid increases with (a) increase in temperature (b) reduction of gas pressure (c) decrease in temperature (d) amount of liquid taken. 7. Arrange the following aqueous solutions in the order of their increasing boiling points (i) 10 4 M NaCl (ii) 10 4 M Urea (iii) 10 3 M MgCl 2 (iv) 10 2 M NaCl (a) (i) < (ii) < (iv) < (iii) (b) (ii) < (i) (iii) < (iv) (c) (ii) < (i) < (iii) < (iv) (d) (iv) < (iii) < (i) (ii). 8. A solution containing g naphthalene (mol. mass 128) in 50 g of carbon tetrachloride yields a boiling point elevation C while a solution of g of an unknown solute in the same weight of sample solvent gives a boiling point elevation of C. The molecular mass of unknown solute is (a) g mol 1 (b) g mol 1 (c) g mol 1 (d) g mol 1

29 72 Objective MHT-CET Chemistry 9. If 0.5 m solution of Ca(NO 3 ) 2 and 0.75 m solution of KOH is taken, then the depression in freezing point is (a) greater in Ca(NO 3 ) 2 because number of ions are greater (b) greater in KOH because concentration is high (c) equal in both and freezing point is less than 0 C because ionic concentration is same (d) equal to 0 C in both because ionic concentration is negligible. 10. CrCl 3 6NH 3 can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of C. Assuming 100% ionisation of the complex and co-ordination number of Cr as six, the complex will be (K f for water 1.86 K kg mol 1 ) (a) [Cr(NH 3 ) 6 ]Cl 3 (b) [Cr(NH 3 ) 5 Cl]Cl 2 (c) [Cr(NH 3 ) 4 Cl 2 ]Cl (d) [Cr(NH 3 ) 3 Cl 3 ]Cl 11. A m aqueous solution of an ionic compound [Co(NH 3 ) 5 (NO 2 )]Cl freezes at C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (K f 1.86 C/m) (a) 3 (b) 4 (c) 1 (d) Pure benzene freezes at 5.3 C. A solution of g of phenylacetic acid (C 6 H 5 CH 2 COOH) in 4.4 g of benzene (K f 5.12 K kg mol 1 ) freezes at 4.47 C. From this observation, one can conclude that (a) phenylacetic acid exists as such in benzene (b) phenylacetic acid undergoes partial ionisation in benzene (c) phenylacetic acid undergoes complete ionisation in benzene (d) phenylacetic acid dimerises in benzene. 13. What is the molarity of H 2 SO 4 solution, that has a density 1.84 g/cc at 35 C and contains 98% H 2 SO 4 by weight? (a) 18.4 M (b) 18 M (c) 4.18 M (d) 8.14 M 14. Which one of the following statements is false? (a) The correct order of osmotic pressure of 0.01 M aqueous solution of each compound is BaCl 2 > KCl > CH 3 COOH > sucrose. (b) The osmotic pressure (p) of a solution is given by the equation, p MRT, where M is the molarity of the solution. (c) (d) Raoult s law states that the vapour pressure of a component over a solution is proportional to its mole fraction. Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression. 15. When 2 g of a non-volatile solute was dissolved in 90 g of benzene the boiling point of benzene is raised by 0.88 K. Which of the following may be the solute? (K b for benzene 2.53 K kg mol 1 ) (a) CO(NH 2 ) 2 (b) C 6 H 12 O 6 (c) NaCl (d) None of these. 16. A solution containing 25.6 g of sulphur dissolved in 1000 g of naphthalene whose melting point is 80.1 C gave a freezing point lowering of 0.68 C. Calculate formula of sulphur (K f for naphthalene 6.8 K m 1 ). (a) S 6 (b) S 4 (c) S 8 (d) S Phenol associates in benzene to a certain extent to form dimer. A solution containing kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is (K f for benzene 5.12 K kg mol 1 ) (a) 73.4 (b) 50.1 (c) 42.3 (d) The order of increasing freezing point of C 2 H 5 OH, Ba 3 (PO 4 ) 2, Na 2 SO 4, KCl and Li 3 PO 4 is (a) Ba 3 (PO 4 ) 2 < Na 2 SO 4 < Li 3 PO 4 < C 2 H 5 OH < KCl (b) Ba 3 (PO 4 ) 2 < C 2 H 5 OH < Li 3 PO 4 < Na 2 SO 4 < KCl (c) C 2 H 5 OH < KCl < Na 2 SO 4 < Ba 3 (PO 4 ) 2 < Li 3 PO 4 (d) Ba 3 (PO 4 ) 2 < Li 3 PO 4 < Na 2 SO 4 < KCl < C 2 H 5 OH 19. Calculate the weight of ethylene glycol (an effective antifreeze) that must be added to 25 litre water to protect its freezing at 24 C. (K f 1.86 C m 1 ) (a) 20 kg (b) kg (c) 200 kg (d) kg g of monofluoroacetic acid was dissolved in 0.5 kg of water. The freezing point of solution was observed to be 1.0 C. The van t Hoff factor, degree of dissociation and dissociation constant of acid are respectively (Atomic masses C 12, H 1, F 19, O 16 and K f 1.86 K kg mol 1 ) (a) 1.076, and (b) 0.076, and (c) , and (d) None of these.

30 Solutions and Colligative Properties g of monochlorobutyric acid was dissolved in 250 g of water. If dissociation constant of the acid is and K f 1.86 K kg mol 1, the depression of the freezing point will be (C 12, Cl 35, H 1, O 16) (a) 0.56 K (b) 0.65 K (c) K (d) K 22. Which of the following solutions is a 1 M solution? (C 12, H 1, O 16, Ca 39.98, Cl 35.5,Na 23) (a) 0.46 g of C 2 H 5 OH in 100 ml of solution (b) g of CaCl 2 in 1000 ml of solution (c) 0.23 g of CH 3 OH in 100 ml of solution (d) 5.85 g of NaCl in 1000 ml of solution 23. A solution contains 25% H 2 O, 25% C 2 H 5 OH and 50% CH 3 COOH by mass. The mole fraction of H 2 O would be (a) 0.25 (b) 2.5 (c) (d) Match the terms given in Column I with the type of solutions given in Column II. Column I Column II A. Soda water 1. A solution of gas in solid B. Sugar solution 2. A solution of gas in gas C. German silver 3. A solution of solid in liquid D. Air 4. A solution of solid in solid E. Hydrogen gas 5. A solution of gas in liquid in palladium 6. A solution of liquid in solid A B C D E (a) (b) (c) (d) Which statement is true for solution of M H 2 SO 4? (a) 2 2 litre of the solution contains 0.20 mole of SO 4 (b) 2 litre of the solution contains mole of H 3 O + (c) 1 litre of the solution contains mole of H 3 O + (d) 2 1 litre of the solution contains 0.04 mole of SO The mass of a non-volatile solute (molar mass 40 g mol 1 ) which should be dissolved in 114 g octane to reduce its vapour pressure to 80% is (a) 20 g (b) 30 g (c) 10 g (d) 15 g 27. The partial pressure of ethane over a solution containing g of ethane is 1 bar. If the solution contains g of ethane, then what shall be the partial pressure of the gas? (a) bar (b) 76.2 bar (c) bar (d) 7.6 bar How many grams of sulphuric acid is to be dissolved to prepare 200 ml aqueous solution having concentration of [H 3 O + ] ions 1 M at 25 C temperature? [H 1, O 16, S 32 g mol 1 ] (a) 4.9 g (b) 19.6 g (c) 9.8 g (d) 0.98 g 29. Henry s law constant for CO 2 in water is Pa at 298 K. Calculate the quantity of CO 2 in 500 ml of soda water when packed under 2.5 atm CO 2 pressure at 298 K. (a) kg (b) g (c) kg (d) g 30. How much ml of a 0.1 M HCl is required to react completely with 1 g mixture of Na 2 CO 3 and NaHCO 3 containing equimolar amounts of the two? (a) ml (b) L (c) ml (d) L g of a binary electrolyte (mol. wt. 100) are dissolved in 500 g of water. The depression in freezing point of the solution is 0.74 C (K f 1.86 K molality 1 ). The degree of the ionisation of the electrolyte is (a) 0% (b) 100% (c) 75% (d) 50% 32. Which of the following aqueous solutions produce the same osmotic pressure? (i) 0.1 M NaCl solution (ii) 0.1 M glucose solution (iii) 0.6 g urea in 100 ml solution (iv) 1.0 g of a non-electrolyte solute (X) in 50 ml solution (molar mass of X 200) (a) (i), (ii), (iii) (b) (ii), (iii), (iv) (c) (i), (ii), (iv) (d) (i), (iii), (iv) 33. A solution containing 30 g of a non-volatile solute exactly in 90 g water has a vapour pressure of 2.8 kpa at 298 K. Further 18 g of water is then added to solution, the new vapour pressure becomes 2.9 kpa. The molecular mass of the solute and the vapour pressure of water at 298 K are respectively (a) 25 g and 253 kpa (b) 2.5 g and 2.53 kpa (c) 23 g and 3.53 kpa (d) 2.3 g and 355 kpa 34. Two elements A and B form compounds having molecular formula AB 2 and AB 4. When dissolved in 20 g of benzene (C 6 H 6 ), 1 g of AB 2 lowers the freezing point by 2.3 K whereas 1.0 g of AB 4 lowers it by 1.3 K. The molar depression constant for benzene is

31 74 Objective MHT-CET Chemistry 5.1 K kg mol 1. The atomic masses of A and B are respectively (a) u and u (b) u and u (c) u and u (d) 20 u and 40 u 35. The depression in the freezing point of water when 10 g of CH 3 CH 2 CHClCOOH is added to 250 g of water is (K a , K f 1.86 K kg mol 1 ) (a) 0.25 C (b) 0.35 C (c) 0.55 C (d) 0.65 C mole each of the following solutes are taken in 5 moles of water. A. NaCl B. K 2 SO 4 C. Na 3 PO 4 D. glucose Assuming 100% ionisation of the electrolyte, relative decrease in vapour pressure will be in the order (a) A < B < C < D (b) D < C < B < A (c) D < A < B < C (d) Equal 37. How much C 2 H 5 OH must be added to 1.0 L of H 2 O, so that solution should not freeze at 4 F? [K f (C 2 H 5 OH) 1.86 C/m] (a) < g (b) > g (c) < 20 g (d) g 38. Assume that 0.1 molal solutions of sodium chloride, barium chloride, sodium phosphate and aluminium sulphate are all 100% dissociated. The solution having the highest boiling point is that of (a) NaCl (b) BaCl 2 (c) Na 3 PO 4 (d) Al 2 (SO 4 ) In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionisation is 0.3. Taking K f for water as 1.85 K kg mol 1, the freezing point of the solution will be nearest to (a) C (b) C (c) C (d) C g of NaCl and 180 g of glucose were separately dissolved in 1000 ml of water. Identify the correct statement regarding the elevation of boiling point of the resulting solutions. (a) NaCl solution will show higher elevation of boiling point. (b) Glucose solution will show higher elevation of boiling point. (c) Both the solutions will show equal elevation of boiling point. (d) The boiling point elevation will be shown by neither of the solutions. 41. Which one of the following pairs of solutions are isotonic? (a) 0.1 M urea and 0.1 M NaCl (b) 0.1 M urea and 0.2 M MgCl 2 (c) 0.1 M NaCl and 0.1 M Na 2 SO 4 (d) 0.1 M Ca(NO 3 ) 2 and 0.1 M Na 2 SO Which of the following statements is false? (a) The value of molal depression constant depends on the nature of solvent. (b) Relative lowering of vapour pressure is a dimensionless quantity. (c) Relative lowering of vapour pressure is equal to mole fraction of solvent. (d) Equimolal solutions of different non-electrolyte solutes would elevate the boiling point to the same extent. 43. The freezing point of equimolal aqueous solution will be highest for (a) C 6 H 5 NH + 3 Cl (b) Ca(NO 3 ) 2 (c) La(NO 3 ) 3 (d) C 6 H 12 O Repeated measurements of boiling points of separate solutions of 2.36 g of mercurous chloride in 100 g of water produced DT b values in the range of to K. The atomic weights of mercury and chlorine are 200 and 35.5 respectively. K b 0.5 K per mol for water. These data suggest that mercurous chloride functions as a/an (a) covalent compound in aqueous medium (b) ionic compound in aqueous medium (c) reducing agent in aqueous medium (d) oxidising agent in aqueous medium. 45. Solutions A, B, C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M BaCl 2 and 0.1 M AlCl 3. Which one of the following pairs is isotonic? (a) A and B (b) B and C (c) A and D (d) A and C 46. A solution containing 0.52 g of KCl in 100 g of water froze at 0.25 C. Calculate the percentage ionisation of the salt. (K f 1.86 K m 1, At. masses : K 39.1, Cl 35.5) (a) 93 (b) 83 (c) 73 (d) The van't Hoff factor for 0.1 M Ba(NO 3 ) 2 solution is The degree of dissociation of the salt at this concentration is (a) 91.3% (b) 87% (c) 100% (d) 74%

32 Solutions and Colligative Properties 48. The vapour pressures of two liquids P and Q are 80 and 60 torr respectively. The total vapour pressure of the solution obtained by mixing 3 moles of P and 2 moles of Q would be (a) 68 torr (b) 140 torr (c) 72 torr (d) 20 torr 49. Ebullioscopic constant of water is 0.52 K kg mol 1. The incorrect statement is (a) 30 g of urea in half a kg water produces 0.26 K elevation in boiling point (b) 60 g of urea in one kg water produces 0.52 K elevation in boiling point (c) 90 g of glucose in half a kg water produces 0.52 K elevation in boiling point (d) 120 g of urea in one kg water produces 1.04 K elevation in boiling point. 50. A sugar solution in water is expected to freeze (a) at 273 K (b) below 273 K (c) above 273 K (d) at 298 K. 51. An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7 g of water freezes at K. If the value of K f for water is 1.86 K/m, what is the molecular mass of the monobasic acid? (a) 50.0 g/mol (b) 46.2 g/mol (c) 55.5 g/mol (d) 25.4 g/mol 52. Which one of the following solutions are isotonic? (a) 3.42 g of sugar in 1000 cm 3 of water and 0.18 g of glucose in 1000 cm 3 of water (b) 3.42 g of sugar in 1000 cm 3 of water and 0.18 g of glucose in 100 cm 3 of water (c) 3.42 g of sugar in 1000 cm 3 of water and g of NaCl in 1000 cm 3 of water (d) 3.42 g of sugar in 1000 cm 3 of water and 5.85 g of NaCl in 100 cm 3 of water 53. Two solutions X and Y are separated by a semipermeable membrane. If the solvent flows from X to Y, then (a) X is more concentrated than Y (b) X is less concentrated than Y (c) both X and Y have the same concentration (d) both X and Y get diluted. 54. Which of the following statements is not correct? (a) Osmotic pressure is directly proportional to molar concentration. (b) Hypertonic solutions have lower concentrations with respect to reference solution. (c) Isotonic solutions have same molar concentration. (d) Osmotic pressure depends upon temperature When 5% solution of sucrose (molar mass 342) is isotonic with 1% solution of a compound A, the molar mass of A is (a) 180 g (b) 68.4 g (c) 18 g (d) 32 g 56. A solution is obtained by mixing 300 g of a 25% solution and 400 g of a 40% solution by mass. Calculate the mass percentage of solute in the resulting solution. (a) (b) (c) (d) Given K b and K f of water are 0.52 and 1.86 K m 1. An aqueous solution freezes at C. What is the boiling point of the solution? (a) 0.52 C (b) C (c) C (d) C 58. Pure water as well as separate equimolal and dilute aqueous solutions of NaCl, K 3 [Fe(CN) 6 ] and K 4 [Fe(CN) 6 ] represented respectively as I, II, III and IV are available. Which of the following statements is correct, assuming 100% dissociation of all solutes? (a) I < II < III < IV is the increasing order of freezing points (b) IV < III < II < I is the increasing order of freezing points (c) IV < III < II < I is the increasing order of boiling points (d) IV < III < II is the increasing order of van t Hoff factors. 59. Solution A contains 1 g of urea in 100 ml of water and solution B contains 2 g of glucose in 100 ml of water. Now, (a) boiling point of solution A will be less than solution B (b) freezing point of solution A will be lower than solution B (c) both will have same freezing point and boiling point (d) osmotic pressure of solution A will be less than solution B. 60. Which of the following statements is correct? (a) Lowering of vapour pressure takes place only in ideal solutions (b) Lowering of vapour pressure does not depend upon the solvent at a given concentration of the solute. (c) Lowering of vapour pressure depends upon the nature of solute (d) Relative lowering of vapour pressure does not depend upon the solvent at a given concentration of the solute.

33 76 Objective MHT-CET Chemistry Competitive Exams If M, W and V represent molar mass of solute, mass of solute and volume of solution in litres respectively, which among the following equations is true? (a) p MWR TV (b) p TMR WV (c) p TWR (d) p TRV VM WM (MH-CET) 2. Molarity is defined as (a) the number of moles of solute dissolved in one dm 3 of the solution (b) the number of moles of solute dissolved in 1 kg of solvent (c) the number of moles of solute dissolved in 1 dm 3 of the solvent (d) the number of moles of solute dissolved in 100 ml of the solvent. (MH-CET) 3. van t Hoff factor of centimolal solution of K 3 [Fe(CN) 6 ] is Calculate the percent dissociation of K 3 [Fe(CN) 6 ]. (a) (b) 0.78 (c) 78 (d) (MH-CET) 4. What is the mole fraction of the solute in a 1.00 m aqueous solution? (a) (b) (c) (d) (AIPMT) 5. The vapour pressure of acetone at 20 C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 C, its vapour pressure was 183 torr. The molar mass (g mol 1 ) of the substance is (a) 128 (b) 488 (c) 32 (d) 64 (JEE Main) What is the molality of a solution containing 200 mg of urea (molar mass 60 g mol 1 ) dissolved in 40 g of water? (a) (b) (c) (d) (MH-CET) 7. Solubility of which among the following substances in water increases slightly with rise in temperature? (a) Potassium bromide (b) Potassium chloride (c) Potassium nitrate (d) Sodium nitrate (MH-CET) 8. Consider separate solutions of M C 2 H 5 OH (aq), M Mg 3 (PO 4 ) 2(aq), M KBr (aq) and M Na 3 PO 4(aq) at 25 C. Which statement is true about these solutions, assuming all salts to be strong electrolytes? (a) M C 2 H 5 OH (aq) has the highest osmotic pressure. (b) They all have the same osmotic pressure. (c) M Mg 3 (PO 4 ) 2(aq) has the highest osmotic pressure. (d) M Na 3 PO 4(aq) has the highest osmotic pressure. (JEE Main) p A and p B are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If x A represents the mole fraction of component A, the total pressure of the solution will be (a) p A + x A (p B p A ) (b) p A + x A (p A p B ) (c) p B + x A (p B p A ) (d) p B + x A (p A p B ) (AIPMT) 10. K f for water is 1.86 K kg mol 1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C 2 H 6 O 2 ) must you add to get the freezing point of the solution lowered to 2.8 C? (a) 93 g (b) 39 g (c) 27 g (d) 72 g (AIEEE) 11. The density of a solution prepared by dissolving 120 g of urea (mol. mass 60 u) in 1000 g of water is 1.15 g/ml. The molarity of this solution is (a) 1.78 M (b) 1.02 M (c) 2.05 M (d) 0.50 M (AIEEE)

34 Solutions and Colligative Properties An aqueous solution of urea containing 18 g urea in 1500 cm 3 of the solution has a density equal to If the molecular weight of urea is 60, the molarity of the solution is (a) (b) (c) (d) (MH-CET) g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be (a) (b) (c) (d) (MH-CET) 14. Mole fraction of the solute in a 1.00 molal aqueous solution is (a) (b) (c) (d) (AIPMT) 15. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K f for water is 1.86 C/m, the freezing point of the solution will be (a) 0.18 C (b) 0.54 C (c) 0.36 C (d) 0.24 C (AIPMT Mains) ml of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be bar. The molar mass of protein will be (R L bar mol 1 K 1 ) (a) g mol 1 (b) g mol 1 (c) g mol 1 (d) g mol 1 (AIPMT Mains) 17. The freezing point depression constant for water is 1.86 C m 1. If 5.00 g Na 2 SO 4 is dissolved in g H 2 O, the freezing point is changed by 3.82 C. Calculate the van t Hoff factor for Na 2 SO 4. (a) 2.05 (b) 2.63 (c) 3.11 (d) (AIPMT) 18. The van t Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively (a) less than one and greater than one (b) less than one and less than one (c) greater than one and less than one (d) greater than one and greater than one. (AIPMT) 19. The degree of dissociation (a) of a weak electrolyte, A x B y is related to van t Hoff factor (i) by the expression i 1 i 1 (a) a (b) a ( x+ y 1) ( x+ y+ 1) ( x+ y 1) (c) a i 1 ( x+ y+ 1) (d) a i 1 (AIEEE) 20. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at 6 C will be : (K f for water 1.86 K kg mol 1, and molar mass of ethylene glycol 62 g mol 1 ) (a) g (b) g (c) g (d) g (AIEEE) 21. A 5.2 molal aqueous solution of methyl alcohol, CH 3 OH, is supplied. What is the mole fraction of methyl alcohol in the solution? (a) (b) (c) (d) (AIEEE)

35 78 Objective MHT-CET Chemistry Hints & Explanations Level (a) 2. (c) 3. (a) 4. (d) 5. (a) 6. (a) 7. (c) 8. (d) 9. (a) 10. (c) : Smoke is an aerosol, with gas (air) as the dispersion medium and solid (soot particles mostly) as the dispersed phase. 11. (c) 12. (b) 13. (a) : Mixture Solute Solvent (a) Chloroform in N 2 gas liquid gas CO 2 in water gas liquid (b) Brass solid solid Camphor in N 2 gas solid gas (c) Sodium amalgam liquid solid Moist air liquid gas (d) H 2 gas in Pd metal gas solid Mixture of N 2 and O 2 gas gas 14. (b) 15. (c) 16. (a) 17. (d) : Precentage by volume of C 2 H 5 OH (v/v) Volume of solute 100 Volume of solution % (d) : M 1 V 1 M 2 V V 2 V 2 2 dm (a) : Molality Moles of solute 1000 g of solvent 20. W (a) : M M 3 M2 V( incm ) n (a) : m W1 in g m 22. (a) W (b) : Molarity M2 V Molarity 04. M (Qdensity of water 1 g/cm 3 ) Number of moles of solute 24. (b) : Molality 1000 Weight of solvent (in g) m m (b) : Moles of C 2 H 5 OH Moles of CH 3 COOH % of H 2 O 100 ( ) 25 \ Moles of H 2 O Total number of moles 2.76 Mole fraction of acetic acid (b) : ppm (b) : Percentage by mass of urea (w/w) mass of urea 100 mass of urea + mass of water 6g g g 186 %by mass 28. (c) : Percentage by volume of ethyl alcohol volume of ethyl alcohol 100 volume of solution 3 58 cm % by volume cm 29. (d) : Moles of ethyl alcohol (n 2 ) mass of ethyl alcohol molar mass of ethyl alcohol 23 g 05. mol 1 46 gmol mass of water Moles of water (n 1 ) molar mass of water 54 g 3.0 mol 1 18 gmol n Mole fraction of ethanol x n1+ n n Mole fraction of water x n1+ n and x 1 + x

36 Solutions and Colligative Properties 30. (c) : 35% (w/w) solution of ethylene glycol in water means it has 35 g of ethylene glycol and 65 g of water. Number of moles of ethylene glycol Mass of ethylene glycol Molar mass of ethylene glycol moles 62 Mass of water Number of moles of water Molar mass of water moles 18 \ Mole fraction of ethylene glycol (c) : Mass of NaOH kg, Volume of solution 500 cm dm 3, Molar mass of NaOH kg mol 1 Molarity of NaOH solution mass of NaOH in kg 1 molar mass of NaOH in kg mol volume of solution in dm kg moldm kg mol 05. dm 32. (b) : Molar mass of urea, NH 2 CONH g mol 1 Mass of solvent 100 g 0.1 kg g Moles of urea mol 1 60 gmol moles of urea Molality of solution mass of solvent in kg mol mol kg 01. kg 33. (c) : Molar mass of sugar, C 12 H 22 O 11 (12 12) + (1 22) + (11 16) 342 g mol kg mol 1 Mass of water in syrup mass of syrup mass of sugar g 34.2 g 180 g kg mass of sugar Number of moles of sugar molar mass of sugar kg 0.1 mol kg mol mass of water Moles of water molar mass of water kg 10 moles kg mol 79 moles of sugar Molality mass of solvent in kg 01. mol mol kg 1 kg Mole fraction of sugar in syrup moles of sugar moles of sugar + moles of water (a) : H 2 SO 4 solution is 27% by mass Hence, if mass of H 2 SO 4 is 27 g Then mass of H 2 O is (100 27) 73 g Molality of H 2 SO 4 mass of H SO 2 4 molar mass of H2SO4 mass of solvent in kg kg 3.77 mol kg kg mol kg Mass of solute Density of solution Volume of solution 100 g Therefore, volume of solution g cm cm dm 3 Molarity of solution Mass of solute Molar mass of solute Volume of solution in dm kg kg mol dm mol dm (c) : HCl solution is 38% by mass. Hence, 100 g HCl solution contains 38 g HCl and (100 38) 62 g water. Volume of 100 g HCl solution Mass of solution Density of solution 100 g gcm cm dm 3 Molarity of solution Mass of HCl 3 Molar mass of HCl Volume of solution in dm kg kg mol dm mol dm 3 Mass of HCl Moles of HCl, n 2 Molar mass of HCl kg kg mol 1.04 mol

37 80 Objective MHT-CET Chemistry Mass of H O Moles of H 2 O, n 1 2 Molar mass of H2O kg 3.44 mol kg mol n Mole fraction of HCl, x n1+ n (b) : 15 ppm (by mass) means 15 g chloroform in 10 6 g of the solution. Mass of solvent 10 6 g Molar mass of CHCl g mol 1 Number of moles of solute Mass of solute Molar mass of solute mol No. of moles of solute \ Molality Mass of solvent in g m (b) : Mass of glucose 34.2 g Mass of water 400 g Percentage by mass of glucose in solution (w/w) Mass of solute Mass of solution % by mass 38. (c) : Mass of water 500 g Percentage by mass of a solute in solution 2.38 Mass of solute Percentage by mass of solute Mass of solution 100 Mass of solute 2.38 Mass of solution 100 Let mass of solute be x. x \ 2.38 x (x + 500) 100x 500 \ 2.38x x x 2.38 x \ x x g mass 39. (b) : Density volume Mass of methyl alcohol g Percentage by mass of methyl alcohol % by mass Number of moles of methyl alcohol Mass of methyl alcohol Molar mass of methyl alcohol moles 32 Number of moles of water Mass of water Molar mass of water moles Mole fraction of methyl alcohol (c) : Volume of benzene 12.8 cm 3 Volume of xylene 16.8 cm 3 Volume of solute % by volume of solute Volume of solution % by volume of benzene % by volume 41. (b) : Solution has 24.8% of HCl, that means 24.8 g of HCl and 75.2 g of H 2 O are present. Number of moles of HCl Mass of HCl moles Molar mass of HCl Number of moles of H 2 O Mass of H2O moles Molar mass of H2O 18 no. of moles of HCl Mole fraction of HCl total no. of moles Mole fraction of HCl (a) : 2 molal aqueous solution is given. It means 2 moles of solute are present in 1000 g of solvent i.e., water. Mass Number of moles of H 2 O Molar mass moles 2 Mole fraction of solute (c) : 12.2% HNO 3 means 12.2 g of HNO 3 and 87.8 g of solvent are present. Mass of HNO No. of moles of HNO 3 Molar mass of HNO moles Molality of HNO 3 Mass of HNO3 Molar mass of HNO3 Mass of solvent in kg mol/kg

38 Solutions and Colligative Properties Mass of solution Density of solution Volume of solution 100 \ Volume of solution cm dm 3 Molarity Mass of solute Molar mass of solute Volume of solution in dm mol/dm (b) : 95.8% by mass sulphuric acid is present. Mass of H SO Number of moles of H 2 SO Molar mass of H2SO mol 98 Mass of H 2 O Mass of H O Number of moles of H 2 O 2 Molar mass of H2O \ Mole fraction of H 2 SO Mass of solution Density of solution Volume of solution Mass of solution Volume of solution Density of solution cm dm 3 Molarity of solution Mass of solute Molar mass of solute Volume of solution in dm M (a) : Density of solution g/cm 3 Mass of solution Density of solution Volume of solution Mass of solution \ Volume of solution Density of solution cm \ Volume of solution dm 3 Molarity of solution Mass of solute Molar mass of solute Volume of solution in dm M Molality of solution Mass of NaOH Molar mass of NaOH Mass of solvent in kg mol kg (c) : The given aqueous solution of H 2 SO 4 is 4.22 M and has density of 1.21 g/cm g/cm g/ml Now, 1000 ml of solution contains 4.22 moles of H 2 SO 4 \ Mass of 1000 ml of H 2 SO 4 solution mole of H 2 SO 4 98 g of H 2 SO 4 \ 4.22 mole of H 2 SO g g of H 2 SO g of H 2 SO 4 solution g of H 2 SO 4 So, 100 g of H 2 SO 4 solution g of H 2 SO 4 \ 100 g of H 2 SO 4 solution contains g of H 2 SO 4 and g H 2 O. \ Molality of H 2 SO 4 Mass of H SO 2 4 Molar mass of H2SO4 Mass of solvent in kg mol/kg (b) 48. (b) 49. (d) : Molality 50. (a) Mass of solute Molar mass of solute mass of solvent in kg mol/kg 1 g/ml 51. (d) : Number of moles of H 2 O 18 g/mol mol/ml moles/l 52. (d) : M M, M M V ml, V 2 25 ml Resulting Molarity MV 1 1+ MV 2 2 V1+ V2 ( ) + ( ) 012. M (a) : Molarity n HPO M V in L (d) : Number of moles of Na 2 SO

39 82 Objective MHT-CET Chemistry Number of moles of solute Molarity Volume of solution in L M (a) : Percentage by volume Volume of solute Volume of solution % 56. (d) : 0.8 moles of solute are present in 1000 ml \ 0.1 mole of solute is present in x ml \ x 125 ml 01. x 8 MV (c) : Resulting molarity 1 1 MV 2 2 V1+ V M (d) : Molecular wt. of Na 2 CO g/mol 1 molar means 1000 ml Na 2 CO 3 solution contains 106 g of Na 2 CO M means 1000 ml Na 2 CO 3 solution contains 26.5 g of Na 2 CO ml of 0.25 M Na 2 CO 3 solution contains g of Na 2 CO (a) : HCl NaOH M 1 V 1 M 2 V 2 M HCl M HCl M (a) : 58.5 g of NaCl in 1 litre gives 1 molar solution g of NaCl in 0.5 litre gives 2 molar solution. \ 5.85 g of NaCl in 0.5 litre gives 0.2 molar solution. MV 1 1+ MV (b) : M mix. +. V1+ V M (a) : Concentration of AgNO 3 is 0.03 g/ml. i.e., 1 ml of the solution contains 0.03 g of AgNO 3. Thus, to prepare 60 ml of the solution ( ) 1.8 g of AgNO 3 is required. 63. (b) : 0.5 moles of CaCl 2 are present in 1000 ml of solution 500 ml of solution contains moles of CaCl moles of CaCl 2 CaCl 2 Ca Cl 1 mole 2 moles \ 500 ml of solution contains moles of Cl ions 64. (c) : x A + x B + x C 1 ( na + nb + nc ) 10. n + n + n 65. (a) A B C 66. (b) : According to Henry s law, S KP 67. (a) S mol dm (d) : According to Henry s law S KP K mol dm 3 atm 1 P 0.22 atm Hence, S mol dm 3 atm atm mol dm (c) : According to Henry s law S KP Given S mol dm 3, P 0.24 atm K S P mol dm 024. atm mol dm 3 atm mol dm 3 atm (c) 71. (d) 72. (c) 73. (b) 74. (c) : Alloy of mercury with other metals is known as amalgam, where mercury is in liquid state. 75. (c) : Stainless steel contains chromium and some nickel. 76. (a) 77. (d) 78. (d) 79. (a) 80. (c) : Relative lowering of vapour pressure is a colligative property, not the vapour pressure. 81. (a) 82. (a) 83. (c) 84. (c) 85. (a) 86. (c) 87. (a) 88. (d) : According to Raoult s law, the relative lowering in vapour pressure of a dilute solution is equal to mole fraction of the solute present in the solution. 89. (c) 90. (b) 91. (b) : P total p A + p B x A p A + x B p B p o 5 B or, o + p 5 B p B 160 torr

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