QuestionSolutions / cryoscopic constant
What will be the value of freezing point depression constant of cyclobutane having freezing point of 280 K and molar enthalpy of fusion of 2.68 kJ mol⁻¹? (Molar mass of cyclobutane = 56.108 g mol⁻¹and R= 8.314 JK⁻¹mol⁻¹)
- (a) 14.63 K kg mol⁻¹
- (b) 16.14 K kg mol⁻¹
- (c) 13.64 K kg mol⁻¹
- (d) 12.64 K kg mol⁻¹
Answer: (c) 13.64 K kg mol⁻¹
Kf = R Tf² M/ΔHfus (M in kg mol⁻¹) = 8.314 × 280² × 0.056108 / 2680 = 13.64 K kg mol⁻¹ → (c). Cyclobutane-type solvents have large Kf (like camphor).