Question

For 10 minute each, at $0^{\circ }\text{C}$, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 4 litre capacity. The resulting pressure in the vessel is 2.8 atm and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknown gas ?

• A. $448{\text{g mol}}^{-1}$
• B. $224{\text{g mol}}^{-1}$
• C. $226{\text{g mol}}^{-1}$
• D. None of these

Answer 1

The correct option is A $448{\text{g mol}}^{-1}$

The rate of effusions through the hole for the gases $N_2$ and X in this case is directly proportional to the number of moles of the respective gases in the mixture.
So, Graham's law equation can be applied in terms of number of moles.
$\frac{n_{N_2}}{n_X}=\sqrt{\frac{M_X}{M_{N_2}}}...\left(i\right)$
$P_T{V}_T=n_{T}RT...\left(ii\right)$
$n_T=\frac{2.8\times 4}{0.0821\times 273}=0.5$
$n_X+n_{N_2}=0.5$
As $n_{N_2}=0.4$ (Given)
so, $n_X=0.1$
From $\left(i\right)$ :
$\frac{0.4}{0.1}=\sqrt{\frac{M_X}{28}};M_X=448{\text{g mol}}^{-1}$