A) \[0.532\]
B) \[0.266\]
C) \[0.25\]
D) \[0.174\]
Correct Answer: B
Solution :
Number of moles of \[PC{{l}_{5}}\] dissociated at equilibrium \[=2\times \frac{40}{100}=0.8\] \[\begin{align} & \underset{\begin{smallmatrix} 2\,mol \\ \\ (2-0.8)mol \end{smallmatrix}}{\mathop{PC{{l}_{5}}}}\,\underset{0.8\,mol}{\mathop{\underset{{}}{\mathop{\underset{0}{\mathop{PC{{l}_{3}}}}\,}}\,}}\,\,\,\,+\,\,\,\underset{0.8\,mol}{\mathop{\underset{{}}{\mathop{\underset{0}{\mathop{C{{l}_{2}}}}\,}}\,}}\,initially \\ & at\,\,equilibrium \\ \end{align}\] \[[PC{{l}_{5}}]=\frac{1.2}{2}=0.6\,M{{L}^{-1}}\] \[[PC{{l}_{5}}]=[C{{l}_{2}}]=\frac{0.8}{2}=0.4\,M{{L}^{-1}}\] \[{{K}_{C}}=\frac{[PC{{l}_{3}}]\,[C{{l}_{2}}]}{[PC{{l}_{5}}]}\] \[=\frac{0.4\times 0.4}{0.6}=0.267\,\,mol/d{{m}^{3}}\]
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