JEE Advanced · Chemistry · 1. Mole Concept

The mole fraction of a solute in a solution is $0.1$ . At $298 K$ , the molarity of this solution is the same as its molality. The density of this solution at $298 K$ is $2.0 g c m^{- 3}$ . The ratio of the molecular weights of the solute and solvent, $(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}})$ , is:

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Answer & Solution

Correct Answer

(C) 9

Step-by-step Solution

Detailed explanation

1

mole solution has

0.1

mole solute and

0.9

mole solvent.
Let

M_{1} =

Molar mass of solute

M_{2} =

Molar mass of solvent
Molality,

m = \frac{0.1}{0.9 M_{2}} \times 1000 \dots . . (1)

Molarity,

M = \frac{0.1}{0.1 M_{1} + 0.9 M_{2}} \times 2 \times 1000 \dots . . (2)

∵ m = M

\Rightarrow \frac{0.1 \times 1000}{0.9 M_{2}} = \frac{200}{0.1 M_{1} + 0.9 M_{2}}

\Rightarrow \frac{M_{1}}{M_{2}} = 9

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The mole fraction of a solute in a solution is 0.1. At 298 K, the molarity of this solution is the same as its molality. The density of this solution at 298 K is 2.0 g cm–3. The ratio of the molecular weights of the solute and solvent, MWsoluteMWsolvent, is:

Step-by-step Solution