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Question

# Vapour density of PCl5 is 104.16 but when heated at 230∘C its vapour density is reduced to 62. The percentage dissociation of PCl5 at this temperature will be:

A
6.8%
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B
68%
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C
46%
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D
64%
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Solution

## The correct option is A 68% PCl5⇋PCl3+Cl2At Initial C 0 0At equilibrium C(1−α) Cα Cα Total moles at equilibrium=C(1−α)+Cα+Cα=C(1+α)For a reaction at equilibrium, vapour density is inversely proportional to the number of moles.Therefore, Total moles at equilibriumInitial total mole=Vapour density initialVapour density at equilibriumor, C(1+α)C=Dd1+α=Ddα=Dd−1here, D=104.16d=62Therefore, α=104.1662−1=0.68 or 68%

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