CHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS Hauser- General Chemistry I revised 8/03/08

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1 CHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS Hauser- General Chemistry I revised 8/03/ The solubility of Cr(NO 3 ) 3 9 H 2 O in water is 208 g per 100 g of water at 15 C. A solution of Cr(NO 3 ) 3 9 H 2 O in water at 35 C is formed by dissolving 324 g in 100 g water. When this solution is slowly cooled to 15 C, no precipitate forms. (a) What term describes this solution? SUPERSATURATED. THIS SOLUTION IS NOT AT EQUILIBRIUM AND HOLDS MORE SOLUTE THAT IT SHOULD AT THIS TEMP. (b) What action might you take to initiate crystallization? Use molecular-level processes to explain how your suggested procedure works. ADD A SEED CRYSTAL. THIS MIGHT PROVIDE AN ALIGNED, TEMPLATE CRYSTAL THAT DISSOLVED SOLUTE CAN ADD TO AS IT STARTS TO CRYSTALLIZE Water and glycerol, CH 2 (OH)CH(OH)CH 2 OH, are miscible in all proportions. What does this mean? SOLUBLE IN VIRTUALLY ALL PROPORTIONS. How do the OH groups of the alcohol molecule contribute to this miscibility? THE "OH" GROUPS OF THE GLYCEROL ALLOWS HYDROGEN-BONDING (MIXING) WITH THE HYDROGEN-BONDING WATER MOLECULES (a) Would you expect stearic acid, CH 3 (CH 2 ) 16 COOH, to be more soluble in water or in carbon tetrachloride? Explain. ALTHOUGH STEARIC ACID HAS A "COOH" PORTION THAT MIGHT HYDROGEN-BOND TO WATER, THE LONG NONPOLAR "TAIL" DOMINATES AND STEARIC ACID WOULD BE MORE SOLUBLE WITH THE CARBON TETRACHLORIDE MOLECULE. CARBON TETRACHLORIDE (CCl 4 ) IS A SYMMETRIC TETRAHEDRAL MOLECULE WITH NO H- BONDING POSSIBLE. 1

2 (b) Which would you expect to be more soluble in water, cyclohexane or dioxane? Explain. DIOXANE IS SLIGHTLY MORE SOLUBLE WITH WATER. DIOXANE, ALTHOUGH NOT CAPABLE OF HYDROGEN-BONDING, DOES PRESENT LONE PAIRS ON ITS OXYGENS THAT COULD ATTRACT THE WATER MOLECULE AS WATER SEEKS TO QUENCH ITS PARTIALLY POSITIVE HYDROGEN ATOMS WITH NEGATIVE CHARACTER Which of the following in each pair is likely to be more soluble in hexane, C 6 H 14 : (explain your answer) HEXANE IS NONPOLAR (NO SIGNIFICANT DIPOLE; NO H-BONDING) (b) benzene (C 6 H 6 ) or glycerol, CH 2 (OH)CH(OH)CH 2 OH BENZENE. BENZENE IS NONPOLAR (NO SIGNIFICANT DIPOLE; NO H- BONDING). SHOULD MIX WITH HEXANE. GLYCEROL WILL NOT ALLOW HEXANE TO BREAK UP THE STRONG GLYCEROL HYDROGEN BONDS. (c) octanoic acid, CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 COOH, or acetic acid, CH 3 COOH. OCTANOIC ACID. THE LONG NONPOLAR TAIL OF OCTANOIC ACID WILL ALLOW IT TO INTERACT WITH THE NONPOLAR HEXANE. THE COMPACT ACETIC ACID MOLECULE WILL MAINTAIN ITS HYDROGEN- BONDING WITH ITSELF AND NOT INTERACT WITH HEXANE (a) Explain why carbonated beverages must be stored in sealed containers. NONPOLAR CO 2 IS NOT VERY SOLUBLE IN POLAR WATER, SO HIGH PRESSURE IS REQUIRED TO FORCE IT INTO SOLUTION. (b) Once the beverage has been opened, why does it maintain more carbonation when refrigerated than at room temperature? GASES ARE LESS SOLUBLE AT HOTTER TEMPS, SO A COOLER BEVERAGE WILL CONTAIN HIGHER LEVELS OF CARBONATION. 2

3 13.37 A solution is made containing 14.6 g of CH 3 OH in 184 g H 2 O. Calculate the mole fraction of CH 3 OH. χ = MOLE FRACTION = MOL SUBSTANCE / TOTAL MOLES Convert grams to moles of each substance g CH 3 OH ( 1 mole CH 3 OH/ g ) = mol CH 3 OH 184 g H 2 O ( 1 mol H 2 O / g ) = mol H 2 O Beware of the denominator! χ CH 3 OH = mol CH 3 OH / ( mol CH 3 OH mol H 2 O) = = (3 SF) 13.9 The figure shows two identical volumetric flasks containing the same solution at two temperatures. (a) Does the molarity of the solution change with the change in temperature? Explain. MOLARITY IS mol of solute / L of solution. SINCE LIQUID VOLUMES CHANGE WITH TEMP., MOLARITY IS IMPACTED BY TEMPERATURE. (b) Does the molality of the solution change with the change in temperature? Explain. MOLALITY IS mol of solute / kg of solvent. MASS IS NOT IMPACTED BY TEMP CHANGES, SO MOLALITY VALUES ARE INDEPENDENT OF TEMPERATURE Calculate the molality of the following solution: (a) 8.66 g benzene (C 6 H 6 ) dissolved in 23.6 g carbon tetrachloride (CCl 4 ) MOLALITY m = mole of solute / kg of solvent Convert grams of benzene solute to moles 8.66 g C 6 H 6 ( 1 mole C 6 H 6 / g) = mol C 6 H 6 Convert grams of solvent to kg. m = mol / [(23.6 g solvent) ( 1kg / 1000 g)] = = 4.70 m 3

4 13.59 Consider two solutions, one formed by adding 10 g of glucose (C 6 H 12 O 6 ) to 1 L of water and the other formed by adding 10 g of sucrose (C 12 H 22 O 11 ) to 1 L of water. Are the vapor pressures over the two solutions the same? Why or why not? VAPOR PRESSURE IS IMPACTED BY THE # OF SOLUTE PARTICLES IN A SOLUTION. SUCROSE HAS A GREATER MOLAR MASS, SO THE SUCROSE SOLUTION HAS LESS PARTICLES (LOWER MOLE COUNT) INTERFERING WITH THE VAPORIZATION PROCESS. THIS SUCROSE SOLUTION WILL HAVE A HIGHER VAPOR PRESSURE (a) Why does a 0.10 m aqueous solution of NaCl have a higher boiling point than a 0.10 m aqueous solution of C 6 H 12 O 6? ALTHOUGH THE LABELS ARE THE SAME, WE MUST TAKE INTO ACCOUNT THE DISSOCIATION OF PARTICLES IN THE SOLUTION. NaCl DISSOCIATES INTO TWO PARTS, EFFECTIVELY MAKING A 0.20 m SOLUTION. COVALENT C 6 H 12 O 6 WILL NOT DISSOCIATE. THE GREATER # OF PARTICLE IN THE SODIUM CHLORIDE SOLUTION WILL CAUSE A GREATER BP ELEVATION. (b) Calculate the boiling point of each solution. ΔT = K b m For H 2 O, K b is 0.51 C/m (given) DO NOT FORGET TO ADD ΔT TO THE BP OF PURE WATER (assume C) For 0.10 m NaCl: (0.51 C/m) (0.20 m) = C = C For 0.10 m C 6 H 12 O 6 : (0.51 C/m) (0.10 m) = C = C List the following aqueous solutions in order of increasing boiling point: m glucose, m LiBr, m Zn(NO 3 ) 2. TAKE INTO ACCOUNT DISSOCIATION! GLUCOSE LiBr Zn(NO 3 ) 2 1 unit X m = m 2 units X m = m 3 units X m = m HIGHER CONCENTRATION LEADS TO GREATER BP ELEVATION. lower BP to higher BP LiBr < glucose < Zn(NO 3 ) 2 4

5 13.99 Calculate the freezing point of a m aqueous solution of K 2 SO 4, ignoring interionic attractions. TAKE INTO ACCOUNT DISSOCIATION! ΔT = K f m For H 2 O, K f is 1.86 C/m (given) m K 2 SO 4 X three ions = m particles DO NOT FORGET TO ADJUST ΔT TO THE FP OF PURE WATER (assume 0.00 C) (1.86 C/m) (0.300 m) = C = C = C is FP of solution What is the osmotic pressure formed by dissolving 44.2 mg of aspirin (C 9 H 8 O 4 ) in L of water at 25 C? Π = MRT Calculate M (MOLARITY) MOLE ASPIRIN = C 9 X = H 8 X 1.01 = 8.08 O 4 X = sum = g/mol MANY CONVERSIONS NEEDED: 44.2 mg ( 1 g / 1000 mg) = g 25 C = K (a 3 SF K value) [ g aspirin (1 mol asp / g)] [ L atm/ mol K] [ K] Π = = [0.358 L] = atm (3 SF) OSMOTIC PRESSURE (Π) 5

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