Question

The rate of diffusion of methane at a given temperature is twice that of a gas $X$. The molar mass (in g mol${}^{-1}$) of $X$ is:

Answer 1

As we know,
$\frac{{\gamma }_{C{H}_4}}{{\gamma }_g}=\sqrt{\frac{M_g}{M_{C{H}_4}}}$
or $2=\sqrt{\frac{M_g}{16}}$
$\therefore M_g=64$ g mol${}^{-1}$