An aqueous solution boils at 100.1 degrees C. What is its freezing point? Given: enthalpy of fusion of water = 80 cal/g, enthalpy of vaporisation of water = 540 cal/g.

0.36 degrees C
-0.36 degrees C
-3.6 degrees C
None of these

Correct answer: -0.36 degrees C

Solution

Since both properties depend on molality, delta_Tf/delta_Tb = Kf/Kb = (Tf/Tb)² * (L_vap/L_fus) = (273/373)² * (540/80) ≈ 3.61. With delta_Tb = 0.1 deg C, delta_Tf ≈ 0.36 deg C. Freezing point = -0.36 degrees C.

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