Question

A solution consists of 200. g of a nonelectrolytic solute dissolved in 500. g of water. It freezes at -9.30°C. What is the molecular weight of the solute?

Answer 1

The Freezing point of pure water is `0 ^0`C

Depression in Freezing point is -9.3 `0 ^0`C

Step 1: Calculate the freezing point depression of the solution.

`Delta`Tf = (Freezing point of pure solvent) - (Freezing point of solution)
(`0 ^0`C) - (-9.3 `0 ^0`C) = 9.3 `0 ^0`C

Step 2 : Calculate the molal concentration of the solution using the freezing point depression.

`Delta` Tf = (Kf) (m)

m = (9.3 `0 ^0`C) / (1.8`6^o`C /molal)
m = 5.0 molal

Step 3: Calculate the molecular weight of the unknown using the molal concentration.

m= 5.0 molal

m = number of moles of solute / mass of solvent (kg)

1. 0 molal = number of moles of solute / 0.500 kg

number of moles of solute = ( 5.0 mol/ kg ) x 0.500 kg

number of moles of solute = 2.5 moles

moles of substance X = mass of X / Molar mass of X

2.5 moles = 200 g / Molar Mass of X

Molar Mass of X = 200 g / 2.5 moles = 80 g/mol