The amount of ${\mathrm{PCI}}_5$ to be taken in a one litre vessel in order to obtain a concentration of 0.1 M of Cl₂ at equilibrium when the ${\mathrm{K}}_{\mathrm{c}}$ for ${\mathrm{PCl}}_5\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{PCl}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$ is 0.0414 M. 

${\mathrm{PCl}}_5\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{PCl}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

a                     0                 0       initial

(a – 0.1)        0.1              0.1    at equilibrium

$K_c=\frac{0.1\times 0.1}{(a-0.1)}, 0.0414=\frac{0.01}{(a-0.1)}$

$0.0414 \mathrm{a} – 0.00414 = 0.01, 0.0414 \mathrm{a} = 0.01414, \mathrm{a}=\frac{0.01414}{0.0414}=0.3415\mathrm{mole}$

0.341 mol of  ${\mathrm{PCI}}_5$ is to be taken in one litre vessel to have 0.1 M ${\mathrm{CI}}_2$ at equilibrium.