Cliffs AP Chemistry, 3rd Edition

8. In order to determine the molecular weight of a particular protein, 0.010 g of the protein
   was dissolved in water to make 2.93 mL of solution. The osmotic pressure was
   determined to be 0.821 torr at 20.0°C. What is the molecular weight of the protein?

A. 3.8 × 103 g/mole

B. 7.6 × 103 g/mole

C. 3.8 × 104 g/mole

D. 7.6 × 104 g/mole

E. None of the above

Answer: D

Begin with the equation

MW = gRT/πV

..

1. 
   

0 010 0 0821

1

293

0 821

1

293

1

1

760

1

g 1000

mole K

liter atm K

torr mL atm

torr

liter

mL

# :

:

= #####

= 7.6 × 104 g/mol

9. A solution of NH 3 dissolved in water is 10.0 m. What is the mole fraction of water in
   the solution?

A. 1.00/1.18

B. 1.00/2.18

C. 0.18/1.00

D. 0.18/10.0

E. 1.18

Answer: A

This problem can be solved using the factor-label method:

..
.

.

kg H O

moles NH

gH O

kg H O

mole H O

gH O

mole H O

mole NH

1

10 0

1000

1

1

18 02

100

0 180

2

3

2

2

2

2

2

##=^3

total number of moles = 0.180 mole NH 3 + 1.00 mole H 2 O= 1.18 moles sol’n

mole fraction of water = 118100 .. mole sol’nmole H O^2

Part II: Specific Topics