Concise Physical Chemistry

c12 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come

PROBLEMS, EXAMPLES, AND EXERCISE 199

Further Comment:Sucrose molecules (or any solute molecules) may “tie up” or hy-

drate (solvate) water molecules, rendering them inactive in any participatory physical

phenomenon like the vapor pressure over the solution, which depends on the number

of solvent molecules in the solution that are free to go into the vapor phase. Other

colligativeproperties (Section 12.8) are influenced in the same way by solvation.

Exercise 12.2

The freezing point of pure benzene is 5.49◦C, and its freezing point constant is

Kf= 5. 07 ◦C. A sample of a crystalline unknown was made up such that it contained

18.7 mg of unknown per 1.000 g of benzene. The freezing point of the resulting

solution was found to be 4.76◦C. What was the molar mass M of the unknown?

Solution 12.2 The freezing point depression was

fT= 5. 49 − 4. 76 = 0. 73 ◦C=Kfm

m=

0. 73

5. 07

= 0 .144 g kg−^1

The unknown was in a concentration of 0.0187 g per 1.000 g benzene, which is

18.7 g kg−^1. If the solution of 18.7 g kg−^1 is 0.144m, then, by direct ratio, we obtain

18. 7

0. 144

=

M

1. 000

M=130 g mol−^1

whereMis the experimental molar mass. The unknown might be naphthalene, C 10 H 10.

Problem 12.1

Exactly 10.00 g of NaCl was added to sufficient water to make up 100.0 g of solution.

(a) What was the weight %?

(b) How many moles of NaCl were present?

(c) What was the molality of the solution?

(d) What is the molarity of the solution?

(e) Assuming a density of 1.071 g cm−^3 at 20◦C for this solution (CRC Handbook

of Chemistry and Physics,2008–2009, 89th ed.), what is its volume?

(f) With this new information, find its molarity.