360 cm³ of CH₄ gas diffused through a porous membrane in 15 minutes. Under similar conditions, 120cm³ of another gas diffused in 10 minutes. Find the molar mass of the gas.

Methane
Rate of diffusion of methane
${\mathrm{r}}_1=\frac{\text{ Volume of methane }}{\text{ Time of diffusion of methane }}$
$=\frac{360{\mathrm{cm}}^3}{15\min }=24{\mathrm{cm}}^3{\min }^{-1}$
Molar mass of ${\mathrm{CH}}_4\left({\mathrm{M}}_1\right)=16{\mathrm{gmol}}^{-1}$
Unknown Gas :
Rate of diffusion of unknown gas
${\mathrm{r}}_2=\frac{\text{ Volume of unknown gas }}{\text{ Time of diffusion }}$
$=\frac{120{\mathrm{cm}}^3}{10min}=12{\mathrm{cm}}^3{\min }^{-1}$
Molar mass of unknown gas (M₂) = ?
According to Graham’s law of diffusion
$\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\sqrt{\frac{{\mathrm{M}}_2}{{\mathrm{M}}_1}}$
$\frac{24{\mathrm{cm}}^3{\min }^{-1}}{12{\mathrm{cm}}^3{\min }^{-1}}=\sqrt{\frac{{\mathrm{M}}_2}{16\mathrm{g}{\mathrm{mol}}^{-1}}}$
Squaring on both sides.
${\mathrm{M}}_2=\frac{24\times 24\times 16}{12\times 12}=64$
Molar mass of the unknown gas = 64g.mol^–1