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1 mol $P C l_{3}$ and 1 mol $C l_{2}$ are taken in a 10 lit vessel & heated. Now, if 0.8 mol $P C l_{5}$ is present at equilibrium, then

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$[P C l_{3}] = 0.8 M$

$[C l_{2}] = 0.02 M$

$[P C l_{5}] = 0.02 M$

$[P C l_{3}] = [C l_{2}] = 0.2 M$

answer is B.

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Detailed Solution

$P C l_{3} + C l_{2} ⇌ P C l_{5}$

$1 1 0$

$\underline{x x x}$

$1 - x 1 - x x$

$0.2 0.2 0.8$

$G i v e n x = 0.8$

$[C l_{2}] = \frac{0.2}{10} = 0.02 M$

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