Answer:
(b)\[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{1.2\times
{{10}^{3}}\times 1.2\times {{10}^{-3}}}{0.8\times {{10}^{-3}}}\]
\[=1.8\times
{{10}^{-3}}mol\,{{L}^{-1}}\]
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