Question

Calculate the degree of dissociation of $PCl _{5}$, the vapour density at $230^{\circ} C$ is $70$ .

• A. $97.8 \%$
• B. $48.9 \%$
• C. $4.89 \%$
• D. $24.45 \%$

Answer 1

Answer: (B)

Degree of dissociation $(\alpha)=\frac{D-d}{(n-1) d}$

when $n=$ gaseous moles produced by 1 mole of reactant

$\underset{\text{1 mol}}{{PCl _{5( g )}}} \rightleftharpoons \underset{\text{1 mol}}{{PCl _{3( g )}}}+ \underset{\text{1 mol}}{{Cl _{2( g )}}}$

i.e., $n=2$

$D$ (vapour density) $=\frac{M \cdot W}{2}=\frac{208.5}{2}=104.25$

$x=\frac{104.25-70}{(2-1) 70}=0.489$

Percentage dissociation $=0.489 \times 100=48: 9 \%$