Question

The rate of diffusion of methane at a given temperature is twice of a gas X. The molar mass of the gas X is:

• A. 64
• B. 32
• C. 16
• D. 8

Answer 1

Answer: (A)

64

Answer 2

Relation between rate of diffusions and molar masses of the gases X and
$C{{H}_{4}}$ is $\frac{{{r}_{x}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{{{M}_{C{{H}_{4}}}}}{{{M}_{x}}}}$
Let the rate of diffusion of gas
$X=a$ $\therefore$
$\frac{a}{2a}=\sqrt{\frac{16}{{{M}_{X}}}}$
( $\because$ $C{{H}_{4}}=12.4=16$ )
$\therefore$ $\frac{1}{2}=\sqrt{\frac{16}{{{M}_{X}}}}$
$\therefore$ ${{M}_{X}}=16\times 4=64$
So, the molar mass of gas X is 64.