1.00 g of a non-electrolyte solute (molar mass 250 g mol⁻¹) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol⁻¹, the freezing point of benzene will be lowered by :

Question

Dr. Abubacker · Accepted Answer

[Solution from source] Molality m = (1.00/250)/(51.2/1000) = 0.004/0.0512 = 0.078 mol kg⁻¹. ΔTf = Kf × m = 5.12 × 0.078 = 0.40 K. Forgetting to convert grams of benzene to kilograms changes the answer by a factor of 1000. Answer (1).

Q. 1.00 g of a non-electrolyte solute (molar mass 250 g mol⁻¹) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol⁻¹, the freezing point of benzene will be lowered by :

Explanation

Reviewed by Dr. Abubacker

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