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The rate of diffusion of two gases A and B are in the ratio 16 : 3. If the ratio of their masses present in the mixture is 2 : 3 then :

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The ratio of their molar masses is 4: 1

The ratio of their moles present ln side the container is 1 : 24

The ratio of their moles present ln side the container is 8 :3

The ratio of their molar masses is 16 :1

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{16}{3} = \frac{2}{3} \times \frac{M_{B}}{M_{A}} \sqrt{\frac{M_{B}}{M_{A}}} \\ {(\frac{M_{B}}{M_{A}})}^{3 / 2} = 8 \\ \frac{M_{B}}{M_{A}} = \frac{4}{1} \end{array}$

Thus, The ratio of their moles present in side the container is 8 :3.

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