Question

Calculate the van't Hoff factor for an aqueous acetic acid solution that has a concentration of 0.400 percent by mass and a freezing point of -0.200^oC?

Answer 1

In colligative property Vant’ hoff factor is important. It is indicated by letter ‘i’.

Here we are given freezing point of the solution so we use following equation in which it relates with depression in freezing point.

Delta Tf = i x m x kf , here i is Van’t Hoff factor, m is molality, kf is freezing point constant.

Lets assume mass of acetic acid solution = 100 g

Mass of acetic acid = 100 g x 0.400 / 100 = 0.400 g

Mass of water = 100-0.400= 99.6 g = Mass of water in kg = 0.0996 kg

Calculation of moles of acetic acid = 0.400 g/ molar mass of acetic acid = 0.400 g / 60.05 g per mol = 0.00666 mol

Molality = mol solute / mass of solvent kg

Delta Tf = freezing point of pure solute – freezing point of solution

= 0⁰C – ( - 0.200) = 0.200

Freezing point constant of water (solvent ) = 1.86 ⁰C /m

Lets plug all these values.

0.200 ⁰C = i x ( 0.00666 / 0.0996 ) x 1.86⁰C / m

i = 1.6

Van’t Hoff factor of acetic acid = 1.60