# Question #60299

Nov 15, 2015

$146 \text{g""/""mol}$

#### Explanation:

Graham's Law states that the rate of effusion is inversely proportional to the square root of the molar mass:

$R \propto \frac{1}{\sqrt{{M}_{r}}} = \frac{k}{\sqrt{{M}_{r}}}$

For 2 gases we can divide to get:

${R}_{1} / {R}_{2} = \sqrt{\frac{{M}_{2}}{{M}_{1}}}$

We can express rate as $\frac{V}{t}$ so:

$\frac{{V}_{1}}{t} / \frac{{V}_{2}}{t} = \sqrt{\frac{{M}_{2}}{{M}_{1}}}$

Since $t$ is the same we can write:

$\therefore \frac{{V}_{1}}{\cancel{t}} . \frac{\cancel{t}}{{V}_{2}} = \sqrt{\frac{{M}_{2}}{{M}_{1}}}$

$\therefore \frac{{V}_{1}}{{V}_{2}} = \sqrt{\frac{{M}_{2}}{{M}_{1}}}$

I'll refer to the unknown gas as gas 1 and assume the ${M}_{r}$ of Argon to be 40 $\Rightarrow$

$\frac{4.83}{9.23} = \sqrt{\frac{40}{{M}_{1}}}$

$\therefore {\left(\frac{4.83}{9.23}\right)}^{2} = \frac{40}{M} _ 1$

$\therefore 0.274 = \frac{40}{M} _ 1$

$\therefore {M}_{1} = \frac{40}{0.274} = 146$