AIPMT PRELIMS 2006 Chemistry Colligative Properties MCQ Question
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:
0.4 K
0.8 K
0.12 K
0.24 K
Correct Answer
Detailed Explanation
The freezing point depression is calculated using ΔTf = iKfm, where m is the molality. Here, m = (1/250) / (51.2/1000) = 0.078 mol/kg. ΔTf = 5.12 × 0.078 = 0.4 K.
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