Question

What is the freezing point of a 50% (volume/volume) aqueous solution of ethylene glycol (Mm ethylene glycol = 62.07 g/mol)? Assume 1 L of solution and that water is the solvent. (d=1.11 g/mL for ethylene glycol and d=1.00 g/mL for water) a. 45.5˚C b. 9.15˚C c. 33.3˚C d. -45.5˚C e. -9.15˚C f. -33.3˚C

Answer 1

When a non volatile solute is added to the solvent then depression in the freezing point is observed. It is one of the colligative property and entirely depends upon number of molecules of solvent in the solution.

It is mathematically given by ,

Where i = vant hoff factor which is 1 for non dissociative moieties like ethylene glycol

m = molality

Wb = amount  of solute in gram = (500mL X 1.11gmL^-1 ) = 555g

M_b = Molar mass of ethylene glycol = 62.07 gmol^-1

W_a = amount of solvent in grams = (500mL X 1.00g/mol) = 500 g

Hence depression in the freezing point is given by,

Hence freezing point decreases by 33.33 units and the freezing point of water is 0^oC

Hence new freezing point is (0-33.33) ^oC = -33.33 ^oC

Hence new freezing point is -33.33 ^oC , choice f is correct