Chapter 15 Review

We covered sections 1-3 in chapter 15. Please read over these sections in you textbook.

·  You will need to know key vocabulary terms and understand how they relate to each other.

·  You must know the 5 formulas used for concentration (% mass, molarity, molality, dilution, and mole fraction)

·  You must know how to solve boiling point elevation and freezing point depression problems. Remember that change in temperature and new boiling/freezing points are different equations.

·  Know what factors effect solvation and effect solubility and how they affect them.

·  Read and decipher solubility curves. See text page 485, problem # 91

Practice problems

1. What is the molarity of a solution formed by mixing 0.20moles of sodium hydroxide with enough water to make a 150mL solution?

0.20 moles NaOH / 1.3 M
0.150 L solution

2. The molality of a solution of chlorine and water is 0.0362 m. The solution contains 3500 grams of water. How much chlorine, in grams, was used to prepare this solution?

0.0362 m / (x) Moles solute
3.500 kg water
x = 0.1267 moles solute / 71 g Cl2 / 8.995 g Cl2
1 mole Cl2

3. How much 6.0M hydrochloric acid would you dilute to make 0.5-L of 3.0M hydrochloric acid?

6.0 M (V1) = (3.0 M) (0.5L)

V1 = 0.25 L

4. How much potassium bromide, in grams, should be added to water to prepare 0.50L of solution with a molarity of 1.25M?

1.25 M / (x) Moles KBr
0.50 L solution
x = 0.625 moles KBr / 119 g KBr / 74.375 g KBr
1 mole KBr

5. What is the molality of a solution containing 0.03g of silver in 4.75g of iron?

0.03 g Ag / 1 mol Ag / 2.8 x 10 -4 mol Ag
106.4 g Ag
2.8 x 10 -4 mol Ag / 0.0591 m
0.00475 kg Fe

6. How much 2.0M sodium hydroxide would you dilute to get 1000.0mL of 0.10M sodium hydroxide?

(2.0M) (V1) = (0.10M) (1L)

V1 = 0.05 L

7. Calculate the percent by mass of 3.55g of sodium chloride dissolved in 88 grams of water.

3.55 g NaCl + 88 g Water = 91.55 g solution

3.55 g NaCl / 100 / 3.9 % NaCl
91.55 g solution
88 g water / 100 / 96.1 % NaCl
91.55 g solution

8. What mass of water must be added to 255.00g of sodium chloride to make a 15.00% by mass aqueous solution?

0.15 / 255.00 g NaCl
(x) g solution

X = 1700 grams solution

1700 gram solution - 255 g NaCl = 1445 grams water needed to make the solution

9. Calculate the mole fraction of a sodium chloride solution if it contains 15.7g of sodium chloride and 100.0g of water.

First using molar mass, convert both substances to moles. Then…

X NaCl / 0.268 mole NaCl / 0.046
5.818 mol sol’n
X H2O / 5.55 mole H2O / 0.95
5.818 mol sol’n

10. Which of the substances shown is probably a gas? How do you know this? Ce2(SO4)3;; it decreases in solubility as temp increases.

11. How many grams of potassium chloride can be dissolved in 300g of water at 50°C? ~130 g

11. At 40°C, 20g of potassium chlorate is dissolved in 100g of water. Is this solution saturated, unsaturated, or supersaturated? supersaturated

13. Which salt is most soluble at 25 °C? CaCl2

14. Which salt is least soluble at 60 °C? Ce2(SO4)3

15. What is the new boiling point if 25g of ethyl alcohol (C2H5OH) is dissolved in 1.0kg of water? What would be the new freezing point of this solution?

ANSWER: 100.278 °C

16. When 5.0g of a solute is added to 25g of water, the new freezing point is -2.5 °C. What is the molecular mass of the unknown compound?

ANSWER: 166.7 g/mol