In the esterification ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{O}\mathrm{H}\left(\mathrm{l}\right)+\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\mathrm{ }\left(\mathrm{l}\right)\rightleftharpoons \mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}{\mathrm{C}}_2{\mathrm{H}}_5\left(\mathrm{l}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction = 0.333. The equilibrium constant is.

Solution

As alcohol and acid are equimolar so the initial concentration would also be same. Suppose the initial concentration is 'C'.

$\frac{\mathrm{x}}{\left(\mathrm{C}-\mathrm{x}\right)+\left(\mathrm{C}-\mathrm{x}\right)+\mathrm{x}+\mathrm{x}}=0.33=\frac{1}{3}$
$3\mathrm{x}=2\mathrm{C}$
$\mathrm{x}=\frac{2\mathrm{C}}{3}$
${\mathrm{K}}_{\mathrm{c}}=\frac{\frac{2\mathrm{C}}{3}\times \frac{2\mathrm{C}}{3}}{\left(\mathrm{C}-\frac{2\mathrm{C}}{3}\right)\times \left(\mathrm{C}-\frac{2\mathrm{C}}{3}\right)}=4$

Hence, the correct answer is 4.