Question

A 8.50-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

This problem is based on colligative property.

We know delta Tf = m x kf

Here delta Tf is depression in freezing point, m is molality, kf is freezing point constant of solvent and here benzene is the solvent.

We first calculate molality and by using given mass of sample in g and molality we can get its molar mas

Solution :

Calculation of delta Tf

Delta Tf = freezing point of the solvent – freezing point of solution

= 5.5 deg C – 1.02 deg C

=4.48 deg C.

Calculation of molality

Molality = delta Tf / kf

= 4.48 deg C / 5.14 deg C per m

= 0.872 m

Calculation of number of moles

Number of moles = molality x mass of solvent in kg

= 0.872 m x 0.250 kg = 0.218 mol

Molar mass = mass in g / number of moles

= 8.50 g / 0.218 mol

= 39.0 g/mol

Molar mass of the unknown sample would be 39.0 g/mol