(B) The dissociation reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Initial moles: $4 \ mol$ of $PCl_5$,$0 \ mol$ of $PCl_3$,$0 \

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(B) The dissociation reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Initial moles: $4 \ mol$ of $PCl_5$,$0 \ mol$ of $PCl_3$,$0 \ mol$ of $Cl_2$.Degree of dissociation $(\alpha)$ = $0.5$.Moles dissociated = $\alpha \times \text{initial moles} = 0.5 \times 4 = 2.0 \ mol$.At equilibrium:$PCl_5 = 4 - 2 = 2 \ mol$$PCl_3 = 2 \ mol$$Cl_2 = 2 \ mol$Total moles at equilibrium = $2 + 2 + 2 = 6 \ mol$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Degree of dissociation and Vapour density

Medium

$4 \ mol$ of $PCl_5$ is heated in a closed vessel at a constant temperature. If the degree of dissociation for $PCl_5$ is $0.5$,what will be the total number of moles at equilibrium?

A
$4.5$
B
$6$
C
$3$
D
$4$

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1 Equilibrium: Chemical Equilibrium

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Degree of dissociation and Vapour density

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$4 \ mol$ of $PCl_5$ is heated in a closed vessel at a constant temperature. If the degree of dissociation for $PCl_5$ is $0.5$,what will be the total number of moles at equilibrium?

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The density of air is $0.00130 \ g/mL$. The vapour density of air will be

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