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Material Type: Assignment; Professor: Watts; Class: GENERAL CHEMISTRY; Subject: Chemistry; University: Jackson State University; Term: Unknown 1989;
Typology: Assignments
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Solutions to Chapter 12 Practice Problems (Watts). Note: these are my solutions. I have worked them out and typed them from scratch, which takes me a lot of time and effort. They have not been copied from the textbook publisher’s materials. Problem 1 Strictly, none of them, I would say. Answer (b) is close: the density of water is about 1 g/ mL and the volume due to the CH 3 OH is a small fraction of the solution volume. Problem 10 No. At any temperature, the mole fraction of the more volatile component in the vapor is greater than its mole fraction in the solution. Conversely, the mole fraction of less volatile component in the vapor is less than its mole fraction in the solution. Problem 11 A colligative property is one that depends on the number of solute particles present, but not the identity of the solute particles. Properties of a solution which are not colligative: density; vapor pressure (except in the case of a non-volatile solute). Chief colligative properties discussed in this chapter: vapor pressure of a solution with a non-volative solute; boiling point elevation; freezing point depression; osmotic pressure. Density of a solution is not a colligative property. It will depend on the molar mass of the solute, so it depends on the identity of the solute. Problem 13 The answer is (a). This contains more moles of the substance than (b) and (c). Problem 15 Osmosis is the process by which solvent molecules move through a semi-permeable membrane from a lower concentration solution to a higher concentration solution. Osmotic pressure is the pressure needed to stop osmosis. Problem 24 (a) Mass % of solute = 100 %^3.^96 % 100 4. 12
(b) Mass of ethanol: m = dV = (0.789 g/mL) 5.00 mL = 3.945 g Mass % = 100 %^7.^31 %
Suppose we have 1 L of a 3.49 M glycerol solution (aqueous). This solution contains 3. mol of glycerol. The MM of glycerol (C 3 H 8 O 3 ) is 92.11 g mol-1. Mass of glycerol = (3.49 mol)(92.11 g mol-1) = 321.46 g Mass of 1 L of solution = (1.073 g/mL) 1000 mL = 1073 g Mass from water = (1073 – 321.46) g = 751.5 g Number of moles of water = 751.5 g / (18.02 g mol-1) = 41.70 mol Mole fraction of water: 0.^923
The boiling point elevation ΔTTb is 0.068 oC. From the Kb for water (0.512 oC/m), we can find the molality of the solution: m C m C K T o o b b (^) 0. 133