SOME BASIC CONCEPTS OF CHEMISTRY

SOME BASIC CONCEPTS OF CHEMISTRY Type I : Law of Conservation of Mass In all physical and chemical changes the total mass of reactants is equal to that of products. 1. What mass of AgNO will react with 5.85 g NaCl to produce 14.5 g AgCl and 8.5 g NaNO, if law of conservation is true? Ans. AgNO + NaCl AgCl + NaNO x 5.85 g 14.5 g 8.5 g Here, x + 5.85 = 14.5 + 8.5 x = 17.0 g 2. If 6. g of NaHCO are added to 15.0 g of CH COOH, the residue is found to be 18.0 g. What is the mass of CO 2 released? Ans. CHCOOH + NaHCO CHCOONa + H2O + CO2 [. gm] Type II : Law of Constant Composition or Definite proportions. 2.16 g of Cu-metal when treated with nitric acid followed by ignition of nitrate gave 2.70 g of copper-oxide, In another experiment 1.15 g of copper oxide upon reduction with hydrogen gave 0.92 g Cu. Show that this illustrates law of definite proportions. Ans. i) % of Cu in CuO = 2.16 100 2.70 = 80 % Oxygen = 20 % ii) % of Cu in 2 nd experiment = 0.92 100 1.15 = 80 % Oxygen = 20 % As % is same in both, law is obeyed. 4. Silver chloride is prepared by, i) Dissolving 0.5 g of silver wire in HNO and adding excess HCl to AgNO formed. The weight of this dried AgCl is 0.66 g. ii) Heating 1 g of silver metal in a current of dry chlorine gas till metal is completely converted to its chloride. It is found to weigh 1.2 g. Show that these data obey law of constant composition. 1

Type III : Law of Multiple proportions 5. Two oxides of metal contain 27.6 % and 0.0 % of oxygen respectively. If the formula of 1 st oxide is M O 4, find that of the second. Ans. 1 st oxide Mass of oxygen : 27.6 Mass of metal : 100-27.6 = 72.4 2 nd oxide Mass of oxygen : 0 Mass of metal : 100-0 = 70 Formula of 1 st oxide : M O 4 No. of atoms of metal in 2 nd oxide = 70 72.4 = 2.9 No. of atoms of oxygen in 2 nd oxide = 4 0 27.6 = 4.5 Ratio of metal : oxygen in 2 nd oxide = 2.9 : 4.5 or 2 : Formula of 2 nd oxide is M 2O. 6. Copper gives two oxides. On heating 7.0 g of each in hydrogen gas 0.888 g and 0.799 g of metal are produced. Show that the results obey law of multiple proportions. 7. A metal forms two oxides. One contains 46.67 % of metal and other contains 6.94 % metal. Show that the results are as per law of multiple proportions. Type IV : Law of Reciprocal proportion 8. Ammonia contains 82.5 % nitrogen and 17.65 % hydrogen. Water contains 88.9% oxygen and 11.1 % hydrogen. Nitrogen trioxide contains 6.15 % oxygen and 6.85 % of nitrogen. Show that these data are as per law of reciprocal proportion. Ans. In NH, 17.65 11.1 NH H 2 O 82.5 1 g of H combines with 82.5 6.85 N O 2 = 4.67 g of nitrogen 17.65 In H 2O, 1 g of H combines with 88.90 = 8.01 g of oxygen 11.1 Ratio of mass of N and O combining with 1 g H = 4.67 : 8.01 or 1 : 1.72 Ratio of N and O in N 2O = 6.85 : 6.15 or 1 : 1.71 88.90 6.15 Ratios are same. 9. CO 2 contains 27.27 % carbon, CS 2 contains 15.79 % of carbon and SO 2 contains 50 % of sulpher. Are these data in agreement with law of reciprocal proportion? 2

10. 61.8 g A combines with 80 g B. 0.9 g of A combine with 106.5 g of C. B and C combine to form compound CB2. Atomic weights of C and B are respectively 5.5 and 6.6. Show that these data are in accordance with law of reciprocal proportion. Type V : Calculating average atomic mass 11. Chlorine has two isotopes of a.m.u. 4.97 and 6.97 respectively. The relative abundances of these isotopes are 0.755 and 0.245 respectively. Find average atomic mass of chlorine. Ans. Average atomic mass = (4.97 0.755) + (6.97 0.245) = 5.46 12. Naturally occurring Boron consists of two isotopes whose atomic masses are 10.01 and 11.01 respectively. The atomic mass of natural Boron is 10.81. Calculate percentage of each isotope in natural boron. Ans. 20 % and 80 % Type VI : Calculating mass of single atom or molecule Molar mass Mass of single atom or molecule = 2 6.022 10 ( ) N A 1. Calculate mass of i) One atom of Ag ii) One molecule of CO 2. Molar mass of Ag 108 Ans. i) Mass of 1 atom of Ag = = 2 ( N A ) 6.022 10 = 1.79 10-22 g Molar mass of CO2 44 ii) Mass of 1 molecule of CO 2 = ( N A ) 6.022 10 = 7.07 10-2 g = 2 Type VII : Calculation of no. of atoms or molecules in a given mass of substance No. of atoms or molecules = No. of moles N A 14. How many atoms or molecules of sulpher are present in 64.0 g sulpher (S 8)? g Ans. Molar mass of sulpher (S 8) = 8 2 = 256 mol 64 No. of moles of S 8 = 256 = 0.25 No. of molecules = 0.25 6.022 10 2 = 1.50 10 2 molecules No. of S-atoms = 1.50 10 2 8 = 1.20 10 24 atoms

15. Calculate no. of molecules present in i) 4.20 g of cane-sugar (C12H22O11) ii) One drop of water having mass 0.05 g Ans. 6.022 10 22 and 1.67 10 21 respectively 16. Calculate no. of atoms of all constituent elements in 5 g Na 2CO. Ans. Na=6.022 10 2 ions; C=.01 10 2 ; O=9.0 10 2 Type VIII : Calculating no. of atoms and molecules in given volume of gas No. of moles of a gas at STP or NTP = Vol. of gas in litres 22.4 17. Calculate the mass of i) 1 10 2 molecules of methane ii) 112 cm (ml) of hydrogen at STP. 18. Calculate the volume occupied at STP by, i) 14 g of Nitrogen ii) 1.5 moles of CO 2 iii) 10 21 molecules of Oxygen Ans. 11.2 L,.6 L, 7.2 ml 19. Calculate no. of moles of each in following i) 10 g of CaCO ii) 1 10 2 molecules of CO2 Ans. 0.1, 0.166 20. Calculate the mass of CO 2 which contains the same no. of molecules as are in 40 g of Oxygen. Ans. 55 g 21. Calculate the mass of Na2CO which will have the same no. of molecules as are in 12. g of MgSO4.7H2O. Ans. 5. g 22. Calculate the volume occupied at STP by, i) 16.0 g of Oxygen ii) 1.5 moles of Oxygen iii) 6.022 10 2 molecules of CO 2 Ans. 11.2 L,.60 L, 22.4 L 2. Calculate the total no. of electrons present in 1.4 g of nitrogen gas. [Hint: each molecule of N2 has 14 electrons] Ans. 4.214 10 2 electrons 4

24. 9.7 10 17 atoms of Fe weigh as much as 1 cc of H2 at STP. What is the atomic mass of iron? Ans. 55.9 25. Chlorophyll contains 2.68 % of magnesium by mass. Calculate the no. of magnesium atoms in 2.0 g chlorophyll. Ans. 1.45 10 21 Type IX : Calculation of Molarity and Normality of solution obtained upon mixing two or more solutions M1V 1 + M2V 2 +... M = V + V +... 1 2 N = N1V 1 + N2V 2 +... V + V +... 1 2 26. Calculate molarities and normalities of the solutions obtained on mixing (i) 100 ml 0.2 M H 2SO 4 + 50 ml 0.1 M HCl M1V 1 + M2V2 (0.2 100) + (50 0.1) Ans. M = = = 0.167 M V 1 + V2 150 N1V 1 + N2V2 (0.4 100) + (50 0.1) N = = = 0. N V 1 + V2 150 [Note: 0.2 M H2SO4 = 0.4 N H2SO4; 0.1 M HCl = 0.1 N HCl] 27. Calculate molarities and normalities of solutions obtained upon mixing (i) 100 ml 0.2 N H 2SO 4 + 50 ml 0.1 N HCl Type X : Calculations based on Normality and Molarity Mass of solute dissolved in grams 1 Normality = Equivalent mass Volume of solution in litres Mass of solute dissolved in grams 1 Molarity = Molecular mass of solute Volume of solution in litres Molarity n = Normality 28. Calculate molarity of water if its density is 1000 kg/m. kg 1000 1000 g g Ans. d = 1000 = = 1 m 1000 1000 ml ml 1 lit. vol. of water has 1000 g No. of moles of water present in 1 lit = 1000 18 = 55.56 M 5

29. A solution of oxalic acid (COOH)2.2H2O is prepared by dissolving 0.6 g of acid in 250 cm of solution. Calculate (i) molarity and (ii) normality. g [Mol. mass of (COOH)2.2H2O = 12.6 ] mol 0.6 1 Ans. Molarity = = 0.02 M 126 0.250 Normality = Molarity n = 0.02 2 = 0.04 N 0. Commercially available conc. HCl contains 8% HCl by mass. Find g molarity.[density=1.19 cm ] Ans. 8% by mass = 8 g HCl in 100 g solution 8 g HCl in 100 ml = 84 ml solution 1.19 8 1 Molarity = = 12.9 M 6.5 0.084 Type XI : If a solution of fixed concentration is diluted then final normality/molarity can be calculated as Initial Conc. Final Conc. = M V M V 1 1 2 2 or N 1 V1 = N 2 V2 1. What volume of 12.8 M HCl is needed to make 1.00 L solution of 0.1 M? Ans. M1V1 = M2V2 0.1 1000 V1 = = 8.07 ml 12.8 2. Concentrated H 2SO 4 is 98% by mass having density of 1.84 g/cc. What volume of H 2SO 4 is needed to make 5.0 litre solution of 0.5 M? Find molarity of H 2SO 4 solution. Ans. 98% = 98 g H2SO4 in 100 g solution 98 g H 2SO 4 in 100 = 54.4 ml solution 1.84 98 1 Molarity = = 18.4 M 98 0.0544 Now, M 1V 1 = M 2V 2 V1 = 0.5 5 18.4 = 0.158 L = 15.8 ml. A sample of NaOH weighing 0.40 g is dissolved in water and the solution is made to 50.0 cm in volumetric flask. What is the molarity of resulting solution? Ans. 0.2 M 6

4. How many moles and how many grams of NaCl are present in 250 cm of a 0.50 M NaCl solution? Ans. 0.125 mole; 7.12 g 5. In a reaction vessel, 0.184 g NaOH is required to be added for completing the reaction. How many millilitres of 0.150 M NaOH solution should be added? Ans. 0.7 ml Type XII : Finding molality, mass-percent and mole fraction Mass of solute dissolved in grams 1 Molality = Mol. mass of solute Mass of solvent in kg Moles of solute Mole fraction of solute = Moles of solute + moles of solvent Mass of solute Mass-percent = 100 Mass of solution 6. A solution is prepared by adding 2 g of substance A to 18 g water. Calculate mass-percent. 2 Ans. Mass-percent = 100 [2 + 18] = 10% 7. Calculate mole-fraction of solute in 1 molal aqueous solution. Ans. 1 molal aqueous solution = 1 mole solute in 1000 g H 2O 1 1 Mole fraction of solute = = 1000 [1 + ] 56.5 = 0.017 18 g 8. The density of M solution of NaCl is 1.25.Calculate molality of solution. ml Ans. molar solution = moles NaCl in 1000 ml solution For molality, mass of solute and mass of solvent are needed. From the given data, Mass of NaCl in solution = 58.5[mol. mass of NaCl] = 175.5 g Mass of solution = volume density = 1000 1.25 = 1250 g Mass of solvent = [Mass of solution-mass of solute] = [1250-175.5] g = 1074.5 g 175.5 1 Molality = = 2.79 m 58.5 1.0745 9. What is the mass percent of solute in the solution obtained by dissolving 5 g of solute in 50 g of water? Ans. 9.1 % 7

Type XIII : Calculations involving acid-solution mixed with a base solution When acid is neutralized by a base then, Equivalents of acid = Equivalents of base i) Finding equivalents in solution of given concentration Vol. of solution in litres Normality = Equivalents ii) Finding equivalents if mass is given in grams Mass of substance given in grams Equivalents = Equivalent mass of substance 40. Calculate normality and molarity of solution obtained by mixing: (a) 100 ml 0.1 M H 2SO 4 + 50 ml 0.1 M NaOH (b) 50 ml 0.1 N H 2SO 4 + 100 ml 0.1 N HCl Ans. (a) Equivalents of 0.1 M H 2SO 4 0.1 M H2SO4 = 0.2 N H2SO4 [n=2] Equivalents = 0.1 0.2 = 0.02 Equivalents of 50 ml 0.1 M NaOH 0.1 M NaOH = 0.1 N NaOH [n=1] Equivalents = 0.05 0.1 = 0.005 Equivalents of H2SO4 = 0.02-0.005 Left unneutralized = 0.015 Equivalents 0.015 Normality of H2SO4 = = = 0.1 Normality = 0.1 N Molarity = 0.1 2 = 0.05 M (b) Same as [a] [0.0 N; 0.0 M] Vol. of solution in litres 0.150 41. Two acids H 2SO 4 and H PO 4 are neutralized separately by same amount of alkali when sulphate and dihydrogen orthophosphate are formed respectively. Find ratio of mass of H 2SO 4 and H PO 4. Ans. 1 : 2 Success seems to be largely a matter of hanging on after others have let go. 8