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When $2 g$ of a gas A is introduced into an evacuated flask kept at $25^{0} C,$ the pressure is found to be 1 atmosphere. If $3 g$ of another gas is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal behavior, the ratio of their molecular weights $M_{A} : M_{B}$ is

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$1 : 3$

$2 : 3$

$3 : 2$

$3 : 1$

answer is B.

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Detailed Solution

$\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}$

$\frac{1}{\frac{2}{M_{A}}} = \frac{0.5}{\frac{3}{M_{B}}}$

$\frac{M_{A}}{M_{B}} = \frac{1}{3}$

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