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 )

Solution

Chemistry ma’am · 3 steps

Turn the volume of formic acid into moles first, then compare the observed drop with the expected one.

1. Volume to moles. One litre of solution holds 0.5 mL of formic acid, which has mass 0.5 × 1.05 = 0.525 g. With M(HCOOH) = 46 g mol⁻¹, that is 0.525/46 = 0.0114 mol.
2. Molality. The solution is very dilute, so 1 L of it is almost exactly 1 kg of water. The molality is therefore m ≈ 0.0114 mol kg⁻¹.

   i = (Δ T_f)/(K_f m) = 0.0405/(1.86 × 0.0114) = 0.0405/0.0212

3. So. i ≈ 1.9, which is option (c). It lies between 1 and 2, as it should for HCOOH ⇌ HCOO⁻ + H⁺, which gives at most 2 particles per molecule.

The idea to keep. When the concentration is given as a volume of solute, use density and molar mass to get moles. Then i is just the observed Δ T_f divided by K_f m.

Answer from the NTA final answer key. Solution by PrepWiser.