CHE2-WA2 - Written Assignment 2 Physical Properties of...

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Written Assignment 2: Physical Properties of Solutions Answer all assigned questions and problems, and show all work. 1. Arrange the following compounds in order of increasing solubility in water: O2, LiCl, Br2, CH3OH (methanol) (4 points) O 2 < Br 2 < LiCl < CH 3 OH (Reference: Chang 12.11) 2. Calculate the percent by mass of the solute in each of the following aqueous solutions: (12 points) 1. 5.50 g of NaBr in 78.2 g of solution 2. 31.0 g of KCl in 152 g of water 3. 4.5 g of toluene in 29 g of benzene 1. S = 5.50 g/78.2 g = 0.0703 = 7.03 %
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2. S = 31.0 g/(31.0 g + 152 g) = 0.169 = 16.9 % 3. It is not an aqueous solution. Pay attention! S = 4.5 g/(4.5 g + 29 g) = 0.13 = 13 % (Reference: Chang 12.15) 4. Calculate the molality of each of the following solutions: (8 points) 1. 14.3 g of sucrose (C12H22O11) in 676 g of water 2. 7.20 moles of ethylene glycol (C2H6O2) in 3546 g of water 1. M = 14.3 g/(342.2965 g/mol * 676 g) = 6.18*10 -5 mol/g = 0.0618 mol/kg 2. M = 7.20 mol/3546 g = 0.00203 mol/g = 2.03 mol/kg (Reference: Chang 12.17) 4. Calculate the molalities of the following aqueous solutions: (12 points) 1. 1.22 M sugar (C12H22O11) solution (density of solution = 1.12 g/mL) 2. 0.87 M NaOH solution (density of solution = 1.04 g/mL) 3. 5.24 M NaHCO3 solution (density of solution = 1.19 g/mL)
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Combination gives: 1. M = 1.22 mol/L/(1000mL/L * 1.12 g/mL – 342.2965 g/mol * 1.22 mol/L) * 1000mL/L = 1.74 mol/kg 2. M = 0.87 mol/L/(1000mL/L * 1.04 g/mL – 39.997 g/mol * 0.87 mol/L) * 1000mL/L = 0.865 mol/kg 0.87 mol/kg 3. M = 5.24 mol/L/(1000mL/L * 1.19 g/mL – 84.007 g/mol * 5.24 mol/L) * 1000mL/L = 6.99 mol/kg (Reference: Chang 12.19) 5. The alcohol content of hard liquor is normally given in terms of the “proof,” which is defined as twice the percentage by volume of ethanol (CH3OH) present.
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