Question

A 6.59-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that of pure benzene. What is the molar mass of this compound? (Note: Kf for benzene = 5.12°C/m.) Ignore significant figures for this problem.

Answer 1

Answer:

30.12 g/m is the molar mass of the compound

Explanation:

Freezing point depression to solve this. The formula for the colligative property is:

ΔT = Kf . m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryoscopic constant

m = molality (mol/kg)

T° freezing pure benzene: 5.5°C

(5.5°C - 1.02°C) = 5.12 °C/m . m

4.48°C = 5.12 °C/m . m

4.48°C / 5.12 °C/m = m → 0.875 mol/kg

Mol = mass / molar mas

Molality = mol /kg

Let's find out the molar mass, with this equation:

(6.59 g / Molar mass)  / 0.250 kg = 0.875 mol/kg

6.59 g / molar mass = 0.875 mol/kg . 0.250 kg

6.59 g / molar mass = 0.21875 mol

6.59 g / 0.21875 mol = molar mass → 30.12 g/m