${\mathrm{PCl}}_{5\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{PCl}}_{3\left(\mathrm{g}\right)}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)},{\mathrm{K}}_{\mathrm{C}}=2.7$

3.0 moles of PCl₅  introduced in a 1L closed reaction vessel at 380K. The number of moles of PCI₅ at equilibrium is………………. $\times {10}^{-3}$M

$\begin{array}{l}{\mathrm{PCl}}_{5\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{PCl}}_{3\left(\mathrm{g}\right)}+{\mathrm{Cl}}_{2\left(\mathrm{g}\right)}\\ \mathrm{At}\mathrm{t}=03\mathrm{moles}00\\ \mathrm{At}\mathrm{t}=\mathrm{\alpha }3-\mathrm{x}\mathrm{x}\mathrm{x}\end{array}$
${\mathrm{k}}_{\mathrm{c}}=2.7\Rightarrow {\mathrm{k}}_{\mathrm{c}}=\frac{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}{\left[{\mathrm{PCl}}_5\right]}$
$2.7=\frac{{\mathrm{x}}^2}{3-\mathrm{x}}$
${\mathrm{x}}^2+2.7\mathrm{x}-8.1=0$
${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$
$\mathrm{a}=1,\mathrm{b}=2.7,\mathrm{c}=-8.1$$\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$
$=\frac{-2.7\pm \sqrt{{\left(2.7\right)}^2+4\left(8.1\right)}}{2}=1.8$
$\mathrm{Moles}\mathrm{of}{\mathrm{PCl}}_5=3-1.8=1.2$
$=1200\times {10}^{-3}$