In a 500 mL capacity vessel CO and ${\mathrm{Cl}}_2$ are mixed to form ${\mathrm{COCl}}_2$.At equilibrium it contains 0.2 mole of  ${\mathrm{COCl}}_2$ and 0.1 mole of each of CO and ${\mathrm{Cl}}_2$.The equilibrium constant ${\mathrm{K}}_{\mathrm{C}}$ for reaction,

$\mathrm{CO}+{\mathrm{Cl}}_2\to {\mathrm{COCl}}_2 \mathrm{is}$

The equilibrium of a reaction is, $\mathrm{CO}+{\mathrm{Cl}}_2\to {\mathrm{COCl}}_2$

Volume of vessel ,V = 500 mL

1000mL=1L

so, 500 mL=$\frac{500}{1000}L=0.5 L$

At equilibrium concentration of species are as follows:

Concentration of CO $=\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.1 \mathrm{mol}}{0.5 \mathrm{L}}=0.2 \mathrm{M}$

Concentration of ${\mathrm{cl}}_2=\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.1 \mathrm{mol}}{0.5 \mathrm{L}}=0.2 \mathrm{M}$

Concentration of  ${\mathrm{COCl}}_2=\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.2 \mathrm{mol}}{0.5 \mathrm{L}}=0.4 \mathrm{M}$

Then, the equilibrium constant is

${\mathrm{K}}_{\mathrm{C}}=\frac{\left[{\mathrm{COCl}}_2\right]}{\left[\mathrm{CO}\right]\left[{\mathrm{Cl}}_2\right]}=\frac{0.4}{0.2\times 0.2}$

Therefore, ${\mathrm{K}}_{\mathrm{c}}=10$