Question

At ${200}^o$C $Pc{l}_5$ dissociates as follows: $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$. It was found that the equilibrium vapours are $62$ times as heavy as hydrogen. The degree of dissociation of $PC{l}_5$ at ${200}^o$C is nearly.

• A. $10\%$
• B. $42\%$
• C. $50\%$
• D. $68\%$

Answer 1

The correct option is D $68\%$

$PC{l}_5⟷PC{l}_3+C{l}_{2}1001-xxx$

Now, $x=\frac{M_f-M_0}{M_0}$ (theoretical molecular mass$=208.5=M_f$)

$x=\frac{208.5-124}{124}=0.6814$

Percentage$⟶68.14\%$