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Spontaneity

The rate of d...

Question

The rate of diffusion of a gas $X$ is twice that of another gas of molecular weight 64. Molecular weight of gas $X$ is:

A

16

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B

32

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C

8

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D

2

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Solution

The correct option is B 16
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$\frac{2 r_{2}}{r_{2}} = \sqrt{\frac{64}{M_{X}}}$
Thus, $M_{X} = 16$

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