Question

Imagine two solutions with the same concentration and the same boiling point, but one has ethanol as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.
Given

Ethanol

normal Boiling point: 78.4

Kb: 1.22

CCL4

normal boiling point: 76.8

Kb: 5.03

Answer 1

Answer: m = 0.42; Tb = 79°

Explanation: The relationship between boiling point of the solvent above a solution is directly proportional to the molal concentration of the solute, i.e.:

ΔT = $K_{b}.m$

where

ΔT is the change in boiling point of the solvent;

$K_{b}$ is the molal boiling point elevation constant;

m is the molal concentration of the solute in the solution;

For there two solutions:

1) Ethanol:

ΔT = $K_{b}.m$

Tb - $T_{normal} = K_{b}.m$

Tb - 78.4 = 1.22.m (1)

2) Carbon Tetrachloride:

Tb - $T_{normal} = K_{b}.m$

Tb - 76.8 = 5.03.m (2)

Solving the system of equations:

Tb - 78.4 = 1.22.m

Tb = 1.22.m + 78.4 (3)

Substituing (3) in (2)

1.22.m + 78.4 - 76.8 = 5.03m

3.81m = 1.6

m = 0.42

With m, find Tb:

T - 76.8 = 5.03.0.42

T = 2.11 + 76.8

T = 79°

Molal concentration is 0.42 and boiling point is 79°