Explanation:
B According to Van der Waal's equation for $\mathrm{n}$ mole of real gas,
$\left(\mathrm{P}+\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$
$\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$
Now, for real gas $\mathrm{O}_{2}$,
We have given,
$\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}}$
Comparing equation (i) and (ii), we get-
$\frac{\mathrm{n}}{\mathrm{V}-\mathrm{nb}}=\frac{1}{2 \mathrm{~V}-\mathrm{b}} \Rightarrow \frac{1}{\mathrm{n}}=2$
Therefore, Number of mole in a container, $\mathrm{n}=\frac{1}{2}$
$\because$ Molar mass of $\mathrm{O}_{2}=32 \mathrm{gm}$
$\therefore$ Mass of gas in the container, $\mathrm{m}=\frac{1}{2} \times 32$
$=16 \mathrm{gm}$