83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ____________ K. (Nearest integer)
[Use : Molal Freezing point depression constant of water = 1.86 K kg mol⁻¹]
Freezing Point of water = 273 K
Atomic masses : C : 12.0 u, O : 16.0 u, H : 1.0 u]

Solution

Chemistry ma’am · 3 steps

Find the molality first, then the drop; the freezing point is 273 K minus that drop.

1. Moles of solute. Ethylene glycol is C₂H₆O₂, so its molar mass is 24 + 6 + 32 = 62 g mol⁻¹. That gives 83/62 = 1.339 mol.
2. Molality. Molality is moles of solute per kilogram of solvent, and 625 g of water is 0.625 kg. So m = 1.339/0.625 = 2.142 mol kg⁻¹.

   Δ T_f = K_f × m = 1.86 × 2.142 = 3.984 K

3. So. Ethylene glycol does not ionise, so i = 1. The freezing point is 273 - 3.984 = 269.016 K, which rounds to 269.

The common slip: 4. Stopping at the depression Δ T_f gives 4, not 269. Subtract it from 273 K to get the freezing point.

The idea to keep. Δ T_f is only the depression. Always subtract it from the pure solvent's freezing point, and keep the solvent mass in kilograms.

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