What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g mol⁻¹ ) and the decrease in freezing point is 0.40 K ?

Solution

Chemistry ma’am · 4 steps

Find the molality first, then Kf is just the freezing point drop divided by it.

1. Δ T_f = K_f m ⇒ K_f = (Δ T_f)/m

2. Moles of solute. n = 1/256 = 3.906 × 10⁻³ mol.
3. Molality. Convert the solvent to kilograms: 50 g = 0.050 kg. So m = (1/256)/0.050 = 1000/(256 × 50) = 0.078125 mol kg⁻¹.
4. Solve for Kf. K_f = 0.40/0.078125 = 5.12 K kg mol⁻¹.
5. So. The freezing point depression constant of the solvent is 5.12 K kg mol⁻¹, which is option (c).

The idea to keep. Molality uses kilograms of solvent, not grams. Once you have m, every colligative constant comes straight from Δ T = K m.

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