Answer:
The given reaction:
\[\underset{At\,equilibrium}{\mathop{{}}}\,\underset{0.05}{\mathop{PC{{l}_{5}}(g)}}\,\rightleftharpoons
\underset{x}{\mathop{PC{{l}_{3}}(g)}}\,+\underset{x}{\mathop{C{{l}_{2}}(g)}}\,\]
\[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\]
\[8.3\times
{{10}^{-3}}=\frac{x\times x}{0.05}\]
\[x=0.02\]
\[[PC{{l}_{3}}]=[C{{l}_{2}}]=0.02M\]
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