${\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3+{\mathrm{Cl}}_2 {\mathrm{K}}_{\mathrm{c}}=1.844$
$3.0$ moles of ${\mathrm{PCl}}_5$ is introduced in a $1\mathrm{L}$ closed reaction vessel at $380 \mathrm{K}.$ The number of moles of ${\mathrm{PCl}}_5$ at equilibrium is $—\times {10}^{-3}$
(Round off to the Nearest Integer)

$$\begin{gathered} {\mathrm{PCl}}_{5( \mathrm{g})}\rightleftharpoons {\mathrm{PCl}}_{3( \mathrm{g})}+{\mathrm{Cl}}_{2( \mathrm{g})} {\mathrm{K}}_{\mathrm{c}}=1.844 \\ \mathrm{t}=0 3 0 0 \\ \mathrm{t}={\mathrm{t}}_{\mathrm{eq}} 3-\mathrm{x} \mathrm{x} \mathrm{x} \end{gathered}$$

$\Rightarrow \frac{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}{\left[{\mathrm{PCl}}_5\right]}=\frac{{\mathrm{x}}^2}{3-\mathrm{x}}=1.844$
$\Rightarrow {\mathrm{x}}^2+1.844-5.532=0$
$\Rightarrow \mathrm{x}=\frac{-1.844+\sqrt{(1.844{)}^2+4\times 5.532}}{2}$
$\equiv 1.60$
$\Rightarrow$ Moles of ${\mathrm{PCl}}_5=3-1.60=1.40$