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Solutions to Practice Problems - General Chemistry | CHEM 142, Assignments of Chemistry

Material Type: Assignment; Professor: Watts; Class: GENERAL CHEMISTRY; Subject: Chemistry; University: Jackson State University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/08/2009

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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 CH3OH 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 =
%96.3%100
12.4100
12.4
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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

  1. 12  

(b) Mass of ethanol: m = dV = (0.789 g/mL) 5.00 mL = 3.945 g Mass % = 100 %^7.^31 %

  1. 0 3. 945
  2. 945   (c) Mass of glycerol: m = dV = (1.324 g/mL) 1.50 mL = 1.986 g Mass of water: m = (0.998 g/mL) 22.25 mL = 22.21 g Mass % = 100 %^8.^21 %
  3. 21 1. 986
  4. 986   Problem 30 This can wait. Problem 36 The solution is 75% H 3 PO 4 by mass. Its density is 1.57 g/mL. Consider a solution that contains 1 kg of H 2 O. If x is the mass of H 3 PO 4 in the solution, then 100 % 75 % 3 kg 1 kg     x x x Hence this sample of solution contains 1 kg of H 2 O and 3 kg of H 3 PO 4. How many moles of H 3 PO 4 (molar mass of H 3 PO 4 is 102.00 g mol-1)? n 29. 41 mol Molalityis 29. 41 m
  5. 00 g mol 3000 g  (^)  1   To find the molarity, first find the volume of the solution in L: Molarity M d m V 11. 5
  6. 548 L
  7. 41 mol 2548 mL 2. 548 L
  8. 57 g/mL 4000 g        Problem 40 Later. Problem 44

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

  1. 70 3. 49
  2. 70 2   xH O  Vapor pressure of water: ^2 ^0.^923 ^17.^5 mmHg^16.^2 mmHg o P xH OP [The 17.5 mmHg is from Table 11.2 (2o oC), page 438]. Problem 58 (a) From Table 12.2 (page 507), the boiling point of benzene is 80.10 oC and Kb for benzene is 2.53 oC m-1. ΔTTb = Kb m = (2.53 oC m-1) 0.44 m = 1.1 oC Tb = ΔTTb + Tbo^ = (1.1 + 80.1) oC = 81.2 oC (b) For water, Tb = 100.00 oC and Kb = 0.512 oC m-1. We must find the molality of the sucrose solution. The molarity is 1.80 M and the density is 1.23 g/mL. Suppose we have a 1 L solution. This contains 1.80 mol of sucrose. The mass is 1230 g. The mass from the sucrose (C 12 H 22 O 11 ; MM = 342.34 g mol-1) is 616.21 g. The mass from the water is therefore 613.8 g = 0.614 kg. Molality = 1.80 mol / 0.614 kg = 2.93 m ΔTTb = Kb m = (0.512 oC m-1) 2.93 m = 1.50 oC Tb = ΔTTb + Tbo^ = (1.50 + 100.00) oC = 101.50 oC Problem 64 Using the methods of Chapter 3, the mass % data tell us that the empirical formula is C 8 H 11 O 4. The empirical mass is 171.19 g mol-1. We will use this information later.

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

  1. 512
  2. 068 m  1    (^)  Now find the number of moles of the solute: n ( 0. 133 m )( 0. 04868 kg ) 6. 47  10 ^3 mol This means that 2.216 g of the solute contains 6.47 x 10-3^ mol. Hence the molar mass of the solute is 343 g mol-1. From the molar mass, the empirical mass, and the empirical formula, we conclude that the molecular formula is C 16 H 22 O 8. Problems 68, 71, 76 : later.