${\mathrm{K}}_{\mathrm{C}}$ for $2{\mathrm{SO}}_2+{\mathrm{O}}_2\rightleftharpoons 2{\mathrm{SO}}_3$ in a 10 lit flask at certain $\mathrm{T}$ is 100 lit-mol${ }^{-1}$. Now, if equilibrium pressures of ${\mathrm{SO}}_2$ and ${\mathrm{SO}}_3$ are equal, then mass of ${\mathrm{O}}_2$ present at equilibrium is

It is given to find the mass of ${\mathrm{O}}_2$ present at equilibrium if equilibrium pressures of ${\mathrm{SO}}_2$ and ${\mathrm{SO}}_3$ are equal. ${\mathrm{K}}_{\mathrm{C}}$ for $2{\mathrm{SO}}_2+{\mathrm{O}}_2\rightleftharpoons 2{\mathrm{SO}}_3$ in a 10 lit flask at certain $\mathrm{T}$ is 100 lit-mol${ }^{-1}$.

It is given that the equilibrium pressures of ${\mathrm{SO}}_2$ and ${\mathrm{SO}}_3$ are equal. Therefore, the number of moles of ${\mathrm{SO}}_2$ and ${\mathrm{SO}}_3$ are also equal at equilibrium.

$\mathrm{a}-2\mathrm{x}=2\mathrm{x}$

$\mathrm{a}=4\mathrm{x}$

$2{\mathrm{SO}}_2+{\mathrm{O}}_2\rightleftharpoons 2{\mathrm{SO}}_3$

${\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\mathrm{SO}}_3\right]}^2}{{\left[{\mathrm{SO}}_2\right]}^2\left[{\mathrm{O}}_2\right]}$

$100=\frac{1}{\left[{\mathrm{O}}_2\right]} \text{ (Since concentration of }{\mathrm{SO}}_3\text{ and }{\mathrm{SO}}_2\text{ are equal) }$

$100=\frac{10}{\mathrm{a}-\mathrm{x}}=\frac{10}{3\mathrm{x}}$

$3\mathrm{x}=0.1$

Number of moles of ${\mathrm{O}}_2$at equilibrium are $\mathrm{a}-\mathrm{x}=3\mathrm{x}=0.1$

Mass of ${\mathrm{O}}_2=0.1\times 32=3.2 \mathrm{g}$

Hence the correct option is (C) $3.2 \mathrm{g}$