The reaction ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+\mathrm{CO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$ is initiated with 1 mol of each of ${\mathrm{H}}_2\mathrm{O}$ and $\mathrm{CO}$ in a flask of volume 10 L. If at equilibrium 40 mass per cent of water reacts, the equilibrium constant of the reaction will be 

Mass of water reacted $=\left(\frac{40\mathrm{g}}{100\mathrm{g}}\times 18\mathrm{g}\right)=7.2\mathrm{g}$  Amount of water reacted $=\frac{7.2\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=0.4\mathrm{mol}$

We will have $\begin{array}{r}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+\mathrm{CO}\left(\mathrm{g}\right)\rightleftharpoons \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\\ (1-0.4)\mathrm{mol}(1-0.4)\mathrm{mol}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.4\mathrm{mol}0.4\mathrm{mol}\end{array}$

Since $\mathrm{\Delta }{v}_{\mathrm{g}}=0$ We will have $K_c=K_n$ Hence

$K_c=\frac{\left(0.4\mathrm{mol}\right)\left(0.4\mathrm{mol}\right)}{\left(0.6\mathrm{mol}\right)\left(0.6\mathrm{mol}\right)}=0.44$