Question

For $10$ min each, at $27{}^{\circ }\text{C}$ from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of $3\text{L}$ capacity. The resulting pressure is $4.18$ bar and the mixture contains $0.4$ mole of nitrogen gas. Thus, molar mass $\left(\text{in g}\right)$ of the unknown gas is:

• A. $417.4$
• B. $28.0$
• C. $14.91$
• D. $2.0$

Answer 1

The correct option is A $417.4$

Given,
The resulting pressure, $P=4.18\text{bar}$
Volume of container, $V=3\text{L}$
Temperature, $T=27{}^{\circ }\text{C}=300\text{K}$
Mole of nitrogen, $n_{N_2}=0.4\text{mol}$
Let mole of gas $=n_g$
$PV=nRT$
$n_{\left(\text{total}\right)}=\frac{PV}{RT}=\frac{4.18\times 3}{0.083\times 300}=0.5036\text{mol}$
$n_g+n_{N_2}=0.5036$
$\therefore n_g=0.5036-0.4000=0.1036\text{mol}$
$r_{N_2}=\frac{n_{N_2}}{t}=\frac{0.4}{10}$
$r_g=\frac{n_g}{t}=\frac{0.1036}{10}$
$\therefore \frac{r_{N_2}}{r_g}=\frac{0.4}{0.1036}=\sqrt{\frac{M_g}{M_{N_2}}}$
$(3.861{)}^2=\frac{M_g}{M_{N_2}}\Rightarrow 14.907=\frac{M_g}{28}$
$M_g=14.907\times 28=417.4\text{g}$