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Solutions question

2006 · Q96
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Solutions question

2006 · Q96

NEETChemistrySolutionsMCQ+4 / −1
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
  1. A
    0.2 K
  2. B
    0.4 K
  3. C
    0.3 K
  4. D
    0.5 K
View written solutionFree

Correct answer: B

Molality of non-electrolyte solute

= 1250×0.0512{1 \over {250 \times 0.0512}}250×0.05121​ = 0.0781 m

Δ\Delta ΔTf = kf m

= 5.12 × 0.0781 = 0.4 K

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