Q. A solution of sucrose (molar mass = 342 g mol⁻¹) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol⁻¹)
A) –0.570°C
B) –0.372°C
C) –0.520°C
D) +0.372°C
Explanation
[Solution from source] Molality = (68.5/342)/1 kg = 0.200 mol/kg. ΔT_f = 1.86 × 0.200 = 0.372 K, so the freezing point is −0.372 °C. +0.372 °C has the wrong sign; freezing point is lowered. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.