Question

Calculate the degree of dissociation of $PC{l}_5$, 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

$\underset{\text{1 mol}}{PC{l}_{5(g)}}\rightleftharpoons \underset{\text{1 mol}}{PC{l}_{3(g)}}+\underset{\text{1 mol}}{C{l}_{2(g)}}$

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