Question

A student determined in the general chemistry lab and aqueous solution made by dissolving 32.44 g of iron three chloride into 500 g of water has a freezing point of -2.12°C what is the Van’t hoff factor for the solution based on his or her experimental data

Answer 1

Ans :

Mass of compound FeCl3 = 32.44 g

moles of compound = mass / molar mass

= 32.44 g / 162.2 g/mol = 0.2 mol

Molality = mol solute / mass of solvent (kg)

= 0.2 mol / 0.500 kg

= 0.4 m

freezing point of water = 0^oC

freezing point depression = freezing point of water - freezing point of solution

= 0^oC - (-2.12^oC)

= 2.12^oC

Freezing point depression = i x kf x m

i is van't hoff factor

kf is freezing constant = 1.86^oC/m

m is molality = 0.4 m

putting values :

2.12^oC = i x 1.86^oC/m x 0.4 m

i = 2.8

So the van't hoff factor will be 2.8