1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to -4°C before freezing. The amount of ice (in g) that will be separated out is __________. (Nearest integer)
[Given : K f (H₂O) = 1.86 K kg mol⁻¹]

Solution

Chemistry ma’am · 3 steps

Only water freezes out, so the sucrose moles stay fixed while the liquid water shrinks.

1. Split the 1 kg. The 1 kg is solution, not water. At 0.75 molal, every 1000 g of water carries 0.75 × 342 = 256.5 g of sucrose, so 1256.5 g of solution holds 1000 g of water. So 1000 g of solution has 1000 × 1000/1256.5 = 795.9 g of water and 0.75 × 0.7959 = 0.597 mol of sucrose.

   Δ T_f = K_f m ⇒ m = 4/1.86 = 2.15 mol kg⁻¹

2. Water still liquid. At -4 °C the liquid must be 2.15 molal. All 0.597 mol of sucrose stays in the liquid, so the liquid water is 0.597/2.15 = 0.2776 kg, that is 277.6 g.
3. So. Ice separated = 795.9 - 277.6 = 518.3 g, which rounds to 518.

The common slip: 651. Taking the 1 kg as pure water gives 0.75 mol of sucrose and 1000 - 348.75 = 651 g of ice. The 1 kg also includes the sucrose.

The idea to keep. When a solution freezes, pure solvent leaves as ice and the solute stays behind. Fix the solute moles first, then use Δ T_f = K_f m to find how much liquid solvent is left.

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