Question

Vapour density of $PC{l}_5$ at equilibrium is 52.125 at $T^{o}C$. Then degree of dissociation of $PC{l}_5$ is: (Mw = 208.5)

• A. 20%
• B. 100%
• C. 30%
• D. 15%

Answer 1

Answer: (B)

100%

Since,

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

Vapour density $=52.125$

$x=\frac{D-d}{(n-1)d}$

$n=$ moles of product obtained by reactants $=2$

$D=\frac{M\left(react\right)}{2}$

$d=\frac{M\left(ons\right)}{2}$

$x=$ degree of dissociation

$x=\frac{D-d}{(2-1)d}=\frac{D}{d}-1$

$x=\frac{\frac{208.5}{2}-52.125}{52.125}$

$x=100\%$

Hence, this is the answer.