5 Steps to a 5 AP Chemistry 2019

Solutions and Colligative Properties ❮ 187

For example, determine the boiling point of a solution prepared by adding 15.00 g
of NaCl to 250.0 g water. (Kb = 0.512 K kg mol-^1 )

∆= =

















































=

=+==°

TiKm −

T

2(0.512Kkgmol )

(15.00gNaCl)

(250.0 g)

1 mole NaCl

58.44 g NaCl

1 kg

1,000 g

1.05 K

(373.15 1.05)K 374.20K( 101.05C)

b

1

bp

A 1.00 molal aqueous solution of trichloroacetic acid (CCl 3 COOH) is heated to the
boiling point. The solution has a boiling point of 100.18°C.
Determine the van’t Hoff factor for trichloroacetic acid (Kb for water =
0.512 K kg mol-^1 ).

DT = (101.18 – 100.00) = 0.18°C = 0.18 K

i = DT/Kbm = 0.18 K/(0.512 K kg mol-^1 )(1.00 mol kg-^1 ) = 0.35

A common mistake is the assumption that the van’t Hoff factor must be a whole
number. This is true only for strong electrolytes at very low concentrations.

Osmotic Pressure

If you were to place a solution and a pure solvent in the same container but separate
them by a semipermeable membrane (which allows the passage of some molecules,
but not all particles) you would observe that the level of the solvent side would decrease
while the solution side would increase. This indicates that the solvent molecules
are passing through the semipermeable membrane, a process called osmosis. Eventually
the system would reach equilibrium, and the difference in levels would remain constant.
The difference in the two levels is related to the osmotic pressure. In fact, one could
exert a pressure on the solution side exceeding the osmotic pressure, and solvent mol-
ecules could be forced back through the semipermeable membrane into the solvent side.
This process is called reverse osmosis and is the basis of the desalination of seawater for
drinking purposes. These processes are shown in Figure 13.1.

Osmotic

pressure

Pure Solution

solvent

Semipermeable

membrane

Net

movement

of solvent

Figure 13.1 Osmotic pressure.