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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2019 · 12 January · Morning Shift · Chemistry · Solutions

The freezing point of a 4% aqueous solution of X is equal to the...

The freezing point of a 44% aqueous solution of XX is equal to the freezing point of a 1212% aqueous solution of YY. If the molecular weight of XX is AA, then the molecular weight of YY will be

Options

  1. A2A2A
  2. BAA
  3. C4A4A
  4. D3A3A

Correct answer

D. 3A3A

Step-by-step solution

Freezing point depression refers to the lowering of the freezing point of solvent upon the addition of solute.

(ΔTf)X=(ΔTf)Y{\left({\Delta T}_{f}\right)}_{X} = {\left({\Delta T}_{f}\right)}_{Y}

kfmx=kfmy{k}_{f}{m}_{x} = {k}_{f}{m}_{y}

4×1000A×96=12×1000M×88\dfrac{4 \times 1000}{A \times 96} = \dfrac{12 \times 1000}{M \times 88}

M=3.27AM = 3 .27 A

≃3A≃ 3A

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