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When 2g of a gaseous substance A is introduced to an initially evacuated flask at 25°C the pressure is found to be 1 atm. 3g of another gaseous substance B is added to it at the same temperature and pressure. The final pressure is found to be 1.5 atm. Assuming ideal gas behaviour, the ratio of molar masses of A and B is

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1:3

3:1

2:3

3:2

answer is A.

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Detailed Solution

According to ideal gas equation PV = nrt

$PV = \frac{g}{m} RT$

Molecular weight of

$M_{A} = \frac{2 RT}{V}$

$M_{B} = \frac{3 \times RT}{0.5 \times V}$

Ratio of molecular masses are

$\begin{array}{l} \frac{M_{A}}{M_{B}} = \frac{2 RT}{V} \times \frac{0.5 V}{3 RT} \end{array}$ = $\frac{2 \times 0.5}{3} = \frac{1}{3}$

Hence the ratio of molecular masses will be 1:3.

Therefore the correct option is A.

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