Vapor density of ${\mathrm{PCl}}_5$ is 104.16 but when heated at $230°\mathrm{C}$ its vapour density is reduced to 62 . The degree of dissociation of ${\mathrm{PCl}}_5$ will be

The vapour density of${\mathrm{PCl}}_5$ is 104.16 but when heated to$230°\mathrm{C}$, its vapour

density is reduced to 62 . The degree of dissociation of ${\mathrm{PCl}}_5$, at$230°\mathrm{C}$ is 0.68

The molecular weight is twice the vapour density.

When the vapor density is 104.16, the molecular weight is 2 times of 104.16=208.3.

When the vapor density is 62, the molecular weight is $2\times 64=124 \mathrm{g}/\mathrm{mol}$$2\times 64=124 \mathrm{g}/\mathrm{mol}$

Let $\mathrm{x}$ be the degree of dissociation of ${\mathrm{PCl}}_5$

               ${\mathrm{PCl}}_5$  ${\mathrm{PCl}}_3$ ${\mathrm{Cl}}_2$

 Initial moles 1 & 0  

 Equilibrium moles  $1-\mathrm{x}$   $\mathrm{\{}x$} $\mathrm{x}$ 

$\frac{208.32(1-\mathrm{x})+137.33\mathrm{x}+71\mathrm{x}}{1-\mathrm{x}+\mathrm{x}+\mathrm{x}}$

$\frac{208.32}{1+\mathrm{x}}=124$

$208.32=124+124\mathrm{x}$ 

$84.32=124\mathrm{x}$ $84.32=124\mathrm{x}$

$\mathrm{x}=0.68$

Hence the correct answer is (B) 68.