JEE Advanced · Chemistry · 1. Mole Concept

A compound $H_{2} X$ with molar weight of 80g is dissolved in a solvent having density of $0.4 g m l^{- 1}$ . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

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Answer & Solution

Correct Answer

(D) 8

Step-by-step Solution

Detailed explanation

Here,

V_{s o l u t i o n} \approx V_{s o l v e n t}

Since, in

1 l

solution, 3.2 moles of solute are present,
So,

1 l

solution

\approx 1 l

solvent (d = 0.4g/ml)

\approx

0.4 kg
So, molality (m)

= \frac{M o l e s o f s o l u t e}{M a s s o f s o l v e n t (k g)} = \frac{3.2}{0.4} = 8

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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A compound H2X with molar weight of 80g is dissolved in a solvent having density of 0.4 gml–1 . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

Step-by-step Solution

A compound $H_{2} X$ with molar weight of 80g is dissolved in a solvent having density of $0.4 g m l^{- 1}$ . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is