Depression Of Freezing Point
When 1.80 g of a non-volatile, non-electrolyte solute is dissolved in 90 g of benzene (Kf = 5.12 K kg mol⁻¹), the freezing point is lowered by 0.80 K. The molar mass of the solute is:
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- depression of freezing point
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
128 g mol⁻¹
Depression of freezing point ΔTf = Kf × m, where m is molality. Rearranging: m = ΔTf / Kf = 0.80 / 5.12 = 0.15625 mol kg⁻¹. Molality = moles of solute / mass of solvent in kg = (W₂/M₂) / (W₁/1000). So 0.15625 = (1.80 / M₂) / (90/1000) = (1.80 / M₂) / 0.09 = 20 / M₂. Therefore M₂ = 20 / 0.15625 = 128 g mol⁻¹. This problem tests the fundamental relationship between freezing point depression and molality, requiring proper unit conversion of solvent mass from grams to kilograms.
This medium difficulty chemistry question is from the chapter solutions, covering the topic of depression of freezing point. It appeared in the 2025 exam.
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