Page 236 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 15] SOLUTIONS 225

15.6. Calculate the molality of a solution prepared by adding 30.0 g of water to a 2.00 m solution containing
50.0 g of water.

2.00 mol solute
Ans. 0.0500 kg H O = 0.100 mol solute
2
1kgH O
2
0.100 mol solute
= 1.25 m
0.0800 kg H O
2
15.7. What mass of solvent is required to make a 2.00 m solution with 3.18 mol of solute?
1 kg solvent

Ans. 3.18 mol solute = 1.59 kg solvent
2.00 mol solute
MOLE FRACTION
15.8. What are the units of mole fraction? Which component is the solvent?
Ans. Mole fraction has no units. It is defined as one number of moles divided by another, and the units cancel.
None of the components is defined as the solvent in dealing with mole fractions.

15.9. Explain why X A = 1.11 cannot be correct.
Ans. The total of all mole fractions in a solution is 1, and no single mole fraction can exceed 1.

15.10. Calculate the molality of C 2 H 5 OH in a solution with mole fraction 0.300 of C 2 H 5 OH in water.
Ans. Assume 1.00 mol total, which contains:
0.300 mol C 2 H 5 OH
18.0gH O

2
0.700 mol H 2 O = 12.6gH O
2
1 mol H 2 O
0.300 mol C 2 H 5 OH
= 23.8 m
0.0126 kg H O
2
15.11. Calculate the molality and the mole fraction of the first compound in each of the following solutions: (a)
1.00 mol CH 2 O and 1.00 mol H 2 O, (b) 0.150 mol CH 3 OH and 50.0 g H 2 O, and (c) 20.0 g C 2 H 5 OH and
50.0 g H 2 O.

Ans. (a) 1.00 mol H 2 O 18.0gH O = 18.0gH O = 0.0180 kg H O
2
1 mol H 2 O 2 2
1.00 mol CH 2 O
Molality = = 55.6 m
0.0180 kg H O
2
1.00 mol CH 2 O
X CH 2 O = = 0.500
1.00 mol CH 2 O + 1.00 mol H 2 O
0.150 mol CH 3 OH
(b) Molality = = 3.00 m
0.0500 kg H O
2
1 mol H 2 O

50.0gH O = 2.78 mol H 2 O
2
18.0gH O
2
0.150 mol CH 3 OH
X CH 3 OH = = 0.0512
2.93 mol total

1 mol C 2 H 5 OH
(c) 20.0gC H 5 OH = 0.435 mol C 2 H 5 OH
2
46.0gC H 5 OH
2
0.435 mol
= 8.70 m
0.0500 kg
From part b we have 2.78 mol H 2 O.
0.435 mol
X C 2 H 5 OH = = 0.135
0.435 mol + 2.78 mol