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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Chemistry
Chapter
Solutions

BITSAT 2019 · Chemistry · Solutions

An aqueous solution freezes at 272. 4K , while pure water freezes...

An aqueous solution freezes at 272.4K272. 4K, while pure water freezes at 273K273K, given Kf=1.86Kkgmol−1{K}_{f} = 1. 86Kkg{mol}^{- 1}, and Kb=0.512Kkgmol−1;{K}_{b} = 0. 512Kkg{mol}^{- 1}; the molality of solution and boiling point of solution respectively will be -

Options

  1. A0.3220. 322 and 373.16K373. 16K
  2. B0.2220. 222 and 273.15K273. 15K
  3. C0.4130. 413 and 400K400K
  4. D0.50. 5 and 300.73K300. 73K

Correct answer

A. 0.3220. 322 and 373.16K373. 16K

Step-by-step solution

Depression in freezing point is given as ΔTf=Kfm{ΔT}_{f} = {K}_{f}m

Molality of solution \left(m\right) = \dfrac{\Delta Tf}{Kf}</p><p>or </p><p>or~m = \dfrac{0 . 6}{1 . 86} = 0 . 322m</p><p>Elevationinboilingpointofsolutionisgivenas</p><p></p><p>Elevation in boiling point of solution is given as</p><p>{ΔT}_{b} = {k}_{b} \times m</p><p></p><p>{ΔT}_{b} = 0 . 512 \times 0 . 322</p><p></p><p>{ΔT}_{b} = 0 . 165</p><p></p><p>∴BoilingpointofsolutionBoiling point of solution= 373 + 0 . 16</p><p></p><p>= 373 . 16K

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