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JEE Main 20199 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2019 · 9 April · Evening Shift · Chemistry · Solutions

Molal depression constant for a solvent is 4.0Kkgmo l ^ - 1 . The...

Molal depression constant for a solvent is 4.0Kkgmol−1.4.0Kkgmo{l}^{- 1} . The depression in the freezing point of the solvent for 0.03molkg−10.03molk{g}^{- 1} solution of K2SO4{K}_{2}S{O}_{4} is:
(Assume complete dissociation of the electrolyte)

Options

  1. A0.36K0.36K
  2. B0.24K0.24K
  3. C0.18K0.18K
  4. D0.12K0.12K

Correct answer

A. 0.36K0.36K

Step-by-step solution

Depression in Freezing Point =ΔTf= \Delta{T}_{f}

ΔTf=i×kf×m\Delta{T}_{f} = i \times {k}_{f} \times m

Molality =0.03mol/kg= 0.03mol / kg

i=3;kf=4Kkg/moli = 3; {k}_{f} = 4Kkg / mol

ΔTf=3×4×0.03=0.36K\Delta{T}_{f} = 3 \times 4 \times 0.03 = 0.36K

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