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The molal freezing point depression constant Kf of a solvent is related to its properties by which expression (R = gas constant, M1 = molar mass of solvent, Tf = freezing point)?
AKf = (1000 delta-fus H) / (R M1 Tf^2)
BKf = (R M1 delta-fus H) / (1000 Tf^2)
CKf = (R Tf^2) / (1000 M1 delta-fus H)
DKf = (R M1 Tf^2) / (1000 delta-fus H)
Answer & Solution
Correct answer: D. Kf = (R M1 Tf^2) / (1000 delta-fus H)
1. The cryoscopic constant depends on the solvent's molar mass, freezing point and enthalpy of fusion.
2. The chapter gives $K_f = (R \times M_1 \times T_f^2)/(1000 \times \Delta_{fus}H)$.
3. So Kf grows with $T_f^2$ and M1, and falls with the enthalpy of fusion.
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.18_
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