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JEE Advanced 2016Paper 1
Exam
JEE Advanced
Year
2016
Paper
Paper 1
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Advanced 2016 · Paper 1 · Chemistry · Some Basic Concepts of Chemistry

The mole fraction of a solute in a solution is 0 . 1 . At 298K ,...

The mole fraction of a solute in a solution is 0.10 . 1. At 298K298K, the molarity of this solution is the same as its molality. The density of this solution at 298K298K is 2.0gcm–32.0gc{m}^{– 3}. The ratio of the molecular weights of the solute and solvent, \left(\dfrac{M{W}_{solute}}{M{W}_{solvent}}\right)$, is:

Correct answer

0

Step-by-step solution

11 mole solution has 0.10 . 1 mole solute and 0.90 . 9 mole solvent.
Let M1={M}_{1} =Molar mass of solute
M2={M}_{2} =Molar mass of solvent
Molality, m=0.10.9M2×1000…..(1)m = \dfrac{0.1}{0.9{M}_{2}} \times 1000 \ldots . . \left(1\right)
Molarity, M=0.10.1M1+0.9M2×2×1000…..(2)M = \dfrac{0.1}{0.1{M}_{1} + 0.9{M}_{2}} \times 2 \times 1000 \ldots . . \left(2\right)
∵m=M∵ m = M
⇒0.1×10000.9M2=2000.1M1+0.9M2⇒ \dfrac{0.1 \times 1000}{0.9{M}_{2}} = \dfrac{200}{0.1{M}_{1} + 0.9{M}_{2}}
⇒M1M2=9⇒ \dfrac{{M}_{1}}{{M}_{2}} = 9
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