STANDARD Chemistry Colligative Properties MCQ Question with Solution

Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y?

A

20

B

50

C

100

D

200

Correct Answer

Option A

Detailed Explanation

Using the formula for relative lowering of vapour pressure: $\frac{P^0 - P_A}{P^0} = \frac{n_B}{n_A}$ , we find the molar mass of Y by substituting the given values: $M_B = \frac{200 \times 2}{20} = 20$ .

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STANDARD Chemistry Colligative Properties MCQ Question

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