The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L^–1 is 0.0405°C. Density of formic acid is 1.05 g mL^–1. The Van't Hoff factor of the formic acid solution is nearly : (Given for water k_f = 1.86 K kg mol^–1)

Complete Solution

Given:

Depression in freezing point, ΔT_f = 0.0405°C

Concentration of formic acid solution = 0.5 mL L^–1

Density of formic acid = 1.05 g mL^–1

Freezing point depression constant for water, k_f = 1.86 K kg mol^–1

Volume of formic acid = 0.5 mL

Density of formic acid = 1.05 g mL^–1

So, mass of formic acid = density × volume:

Mass of formic acid = 1.05 × 0.5 = 0.525 g

Calculate moles of formic acid:

Molar mass of formic acid (HCOOH) = 1 + 12 + (16 × 2) + 1 = 46 g mol^–1

Moles of formic acid = mass / molar mass = 0.525 / 46 ≈ 0.0114 mol

Calculate molality (m):

Since the solution is based on 1 L of water (assumed density of water as 1 kg/L), the mass of solvent is approximately 1 kg.

Molality m ≈ 0.0114 mol/kg

The formula for freezing point depression is:

ΔT_f = i × k_f × m

Substitute the values we know:

0.0405 = i × 1.86 × 0.0114

i = 0.0405 / (1.86 × 0.0114) ≈ 0.0405 / 0.0212 ≈ 1.91

Final Answer:

The Van't Hoff factor (i) of the formic acid solution is approximately 1.91, indicating partial dissociation in solution.