Chapter 12 Solutions

# Chapter 12 Solutions - There are a number of ways to...

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There are a number of ways to express concentration. You have seen Molarity Mole fraction Mass percentage For expressing the composition of a compound Can be used for solutions, as well
All concentration units are fractions The numerator contains the quantity of solute The denominator is the quantity of either solution or solvent They differ in the units used to express these two quantities

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= moles of solute Liters of solution M Molarity (Chapter 4) mol A mol A + mol B + mol C + χ A = ... Mole fraction (Chapter 6) Good for problems involving partial pressures
mass percent = x 100 % grams solute grams solution

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A solution is prepared by dissolving 3.00 g of NaCl in 150 g of water. Express its concentration as mass percent.
molality = mol solute kg solvent Good for experiments where there is a change in temperature and where, consequently, the volume would not stay constant

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What is the molality of a solution prepared by dissolving 3.00 g NaCl in 150 g of water?
Concentration Unit Numerator Units Denominator Units Mass % Grams 100 g solution Molarity Moles 1 L solution Molality Moles 1 kg solvent Mole Fraction Moles 1 mol total of solution

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Express the concentration of a 3.0% H 2 O 2 solution as (a) molality (b) mole fraction
Calculate (a) the molality and (b) the mole fraction of alcohol (C 2 H 5 OH ), in a wine that has an alcohol concentration of 7.5 mass percent.

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Calculate (a) the molality and (b) the mole fraction of alcohol (C 2 H 5 OH ), in a wine that has an alcohol concentration of 7.50 mass percent. Answer: (a) 1.76 m (b) 0.0307
Conversion of most concentration units to molarity usually involve the density of the solution The density of a 12.0% sulfuric acid (H 2 SO 4 , molar mass = 98.08 g/mol) solution is 1.080 g/mL. What is the molarity of this solution?

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For most substances there is a limit to the quantity of solute that dissolves in a specific quantity of the solvent. A dynamic equilibrium exists between the solute particles in solution, and undissolved solute. Sugar (s) Sugar (aq)

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Steps 1 and 2 are endothermic; step 3 is exothermic

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Δ H soln = Δ 1 + Δ 2 + Δ 3 The enthalpy of solution may be either negative (left) or positive (right)
A decrease in enthalpy is an important factor in causing spontaneous change However, many endothermic processes are spontaneous, suggesting another cause of spontaneity. An increase in disorder also favors spontaneous change

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An example of increasing disorder as a driving force is illustrated by the mixing of gases.
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## This note was uploaded on 02/24/2011 for the course CHEM 112 taught by Professor Chen during the Fall '09 term at South Carolina.

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Chapter 12 Solutions - There are a number of ways to...

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