A 7.27-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:. a. 36.5 g/mol
b. 146 g/mol
c. 292 g/mol
d. 5.79 g/mol
e. 73.0 g/mol

146 g/mol → option b.

Explanation:

This is a problem about the freezing point depression. The formula for this colligative property is:

ΔT = Kf . m . i

We assume i = 1, so our compound is not electrolytic.

ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C

m  = molality (mol of solute/kg of solvent)

We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.

We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1

1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg

0.19922 mol/kg = mol/ 0.250kg

mol = 0.19922 . 0.250kg → 0.0498 mol

molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol

it is called acid bonding