40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is __________ K. (Nearest integer) [Given : K f = 1.86 K kg mol⁻¹; Density of water = 1.00 g cm⁻³; Freezing point of water = 273.15 K]

Solution

Chemistry ma’am · 3 steps

Find the molality first, then subtract the depression from 273.15 K.

1. Moles and solvent mass. Glucose: 40/180 = 0.222 mol. With a density of 1.00 g cm⁻³, 200 mL of water weighs 200 g, which is 0.2 kg.
2. Molality. m = 0.222/0.2 = 1.11 mol kg⁻¹. Glucose does not ionise, so i = 1.

   Δ T_f = K_f m = 1.86 × 1.11 = 2.07 K

3. So. T_f = 273.15 - 2.07 = 271.08 K, which rounds to 271.

The common slip: 275. Adding Δ T_f to 273.15 K instead of subtracting it gives about 275.22 K, which rounds to 275, not 271. A solute always lowers the freezing point.

The idea to keep. Molality uses kilograms of solvent, not grams and not litres of solution. Δ T_f is only the drop, so always subtract it from the pure solvent's freezing point.

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