Chemistry II MM of Compound

Question

A 0.520 g sample of an unknown nonelectrolyte compound is dissolved in 4.02 g of lauric acid (Kf = 3.90 °C/m). The freezing point depression is determine to be 4.20 °C. What is the molar mass of the compound?

Answer 1

Answer is 120. 1 g/mol

First, change in F.P = K _f X molality, to find the molality divide the change in freezing point by the K_f

molality = 4.20/3.90 = 1.08 is the molality

Molality = mole of the solute/Kg of the solvent

1.08 = moles of the solute/ 0.00402 kg ( 4.02 grams of the solvent is divided by 1000 to get kilograms)

To get the moles of the solute multiply the molality with the kg of the solvent

= 1.08 X 0.00402 kg = 0.00433 is the moles of the compound

To find the molar mass of the compound divide grams by the number of moles

0.520g/ 0.00433 = 120.1 g/mol