5 Steps to a 5 AP Chemistry

Let’s learn to apply the preceding equation. Determine the freezing point of an aque-

ous solution containing 10.50 g of magnesium bromide in 200.0 g of water.

The most common mistake is to forget to subtract the ΔTvalue from the normal freezing

point.

The freezing-point depression technique is also commonly used to calculate the molar

mass of a solute.

For example, a solution is prepared by dissolving 0.490 g of an unknown compound in

50.00 mL of water. The freezing point of the solution is –0.201°C. Assuming the com-

pound is a nonelectrolyte, what is the molecular mass of the compound? Use 1.00 g/mL as

the density of water.

m=ΔT/Kf=0.201 K/(1.86 K kg mol−^1 ) =0.108 mol/kg

50.00 mL (1.00 g/mL) (1 kg/1000 g) =0.0500 kg

(0.108 mol/kg) (0.0500 kg) =0.00540 mol

0.490 g/0.00540 mol =90.7 g/mol

Many students make the mistake of stopping before they complete this problem.

Boiling-Point Elevation

Just as the freezing point of a solution of a nonvolatile solute is always lower than that of

the pure solvent, the boiling point of a solution is always higher than the solvent’s. Again,

only the number of solute particles affects the boiling point. The mathematical relationship

is similar to the one for the freezing-point depression above and is

ΔTb=iKbmolality

where ΔTbis the number of degrees the boiling point has been elevated (the difference

between the boiling point of the pure solvent and the solution); Kbis the boiling-point ele-

vation constant; the molalityis the molality of the solute; and iis the van’t Hoff factor. You

can calculate a solution’s boiling point if you know the molality of the solution. If you know

the amount of the boiling-point elevation and the molality of the solution, you can calcu-

late the value of the van’t Hoff factor, i.

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 )

Δ= =TiKm 1 3(1.86K kg mol−^1 )

(1mole Mg Br

(gMgBr)1

2

2

)

10 50. 8 84.113g MgBr

g)

2

(.200 0

1

100

kg

0 0g

⎛

⎝

⎜

⎞

⎠

⎟

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ = =

1.59K

Tfp (27 3 3.15 1.59)K 271.56K (−−==°1.59 C)

Solutions and Colligative Properties  185

KEY IDEA

STRATEGY