$0.5$ mole of $\mathrm{PCl_5(g)}$ is placed in a $1\,\text{L}$ vessel at a certain temperature. At equilibrium, $0.15$ mole has dissociated into $\mathrm{PCl_3(g)}$ and $\mathrm{Cl_2(g)}$. The value of $K_c$ for the reaction $\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$ is closest to

1. Initial moles in $1\,\text{L}$ → initial concentrations: $[\mathrm{PCl_5}]_0 = 0.5\,\text{M}$, $[\mathrm{PCl_3}]_0 = [\mathrm{Cl_2}]_0 = 0$. 2. Dissociated amount: $0.15$ mol of PCl$_5$ broke down, so at equilibrium $[\mathrm{PCl_5}] = 0.5 - 0.15 = 0.35\,\text{M}$. 3. Each dissociation event produces 1 PCl$_3$ and 1 Cl$_2$, so $[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = 0.15\,\text{M}$. 4. Apply the expression $K_c = \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \dfrac{(0.15)(0.15)}{0.35}$. 5. Compute: $K_c = \dfrac{0.0225}{0.35} = 0.0643\,\text{M}$, i.e. roughly $0.064\,\text{M}$. 6. Other options come from common slips: B uses $(0.15)^2/(0.5)^2$ in the wrong form; C is the equilibrium $[\mathrm{Cl_2}]$ itself, not $K_c$; D uses $2 \times [\mathrm{Cl_2}]$. _Source: NCERT Class 11 Chemistry, Ch 6, §6.4 (worked-problem method) + §6.4.1 Problem 6.3/6.4 (same algebra), p. 9–11._