Question

Two moles of a nonelectrolyte solute are dissolved in 1 kg of an unknown solvent. the solution freezes at 7.8°c below its normal freezing point. what is the molal freezing-point constant of the unknown solvent? suggest a possible identity of the solvent.

Answer 1

Answer:

3.9°C/m is the molal freezing-point constant of the unknown solvent.

Explanation:

$\Delta T_f=i\timesK_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

i = van't Hoff factor

$K_f$ = freezing point constant

m = molality

we have :

$K_f$ =?

$m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

$m=\frac{2 mol}{1 kg}=2 m$

i = 1 ( non electrolyte)

$\Delta T_f=7.8^oC$

$7.8^oC=1\times K_f\times 2 m$

$K_f=3.9 ^oC/m$

3.9°C/m is the molal freezing-point constant of the unknown solvent.

Answer 2

The solution has a freezing constant of 3.9 °C/m. The solvent is most likely acetic acid.

Using the relation;

ΔT = K m i

ΔT = 7.8°C

m = number of moles of solute/ Mass of solution 2 moles/1 Kg = 2 m

i = Non - electrolyte solution = 1

7.8°C = K. 2 m. 1

K = 7.8°C/2 m. 1

K = 3.9 °C/m

The solvent is most likely acetic acid.

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