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At same temperature and pressure, the rate of diffusion of gas ' $X$ ' is $3 \sqrt{3}$ times that of a gaseous hydrocarbon of molar mass $54 \mathrm{~g} \mathrm{~mol}^{-1}$. The molar mass of $X$ in $\mathrm{g} \mathrm{mol}^{-1}$ is
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According to Graham's law of diffusion rate of diffusion of a gas is inversely proportional to square root of its density.
$$
\text { Rate } \propto \frac{1}{\sqrt{\text { Density }}}
$$
But, density of the gas is proportional to its molar mass, thus
Rate of diffusion $\propto \frac{1}{\sqrt{M_m}}$
For two gases we have
$$
\begin{aligned}
\frac{\text { Rate }_1}{\text { Rate }_2} & =\sqrt{\frac{M_{m_2}}{M_{m_1}}} \\
\left(\frac{\text { Rate }_1}{\text { Rate }_2}\right)^2 & =\frac{M_{m_2}}{M_{m_1}} \\
M_{m_2} & =\left(\frac{\text { Rate }_1}{\text { Rate }_2}\right)^2 M_{m_1} \\
& =\left(\frac{1}{3 \sqrt{3}}\right)^2 \times 54 \\
& =\frac{1}{9 \times 3} \times 54 \\
& =\frac{54}{27}=2 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$
$$
\text { Rate } \propto \frac{1}{\sqrt{\text { Density }}}
$$
But, density of the gas is proportional to its molar mass, thus
Rate of diffusion $\propto \frac{1}{\sqrt{M_m}}$
For two gases we have
$$
\begin{aligned}
\frac{\text { Rate }_1}{\text { Rate }_2} & =\sqrt{\frac{M_{m_2}}{M_{m_1}}} \\
\left(\frac{\text { Rate }_1}{\text { Rate }_2}\right)^2 & =\frac{M_{m_2}}{M_{m_1}} \\
M_{m_2} & =\left(\frac{\text { Rate }_1}{\text { Rate }_2}\right)^2 M_{m_1} \\
& =\left(\frac{1}{3 \sqrt{3}}\right)^2 \times 54 \\
& =\frac{1}{9 \times 3} \times 54 \\
& =\frac{54}{27}=2 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$
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