Question

At $473$ K, partially dissociated vapours of $PC{l}_5$ are $62$ times as heavy as $H_2$.

Calculate the percentage of dissociation of $PC{l}_5$.

• A. $68.15$%
• B. $6.815$%
• C. $34$%
• D. $3.4$%

Answer 1

The correct option is A $68.15$%

The relationship between molar mass and vapour density is $Molarmass=2\times vapourdensity=2\times 62=124$.

	$PC{l}_5$	$PC{l}_3$	$C{l}_2$
Molar mass	208.5	137.5	71
Initial moles	1	0	0
Moles at equilibrium	$1-\alpha$	$\alpha$	$\alpha$
Mole fraction	$\frac{1-\alpha }{1+\alpha }$	$\frac{\alpha }{1+\alpha }$	$\frac{\alpha }{1+\alpha }$

Thus, $208.5\left(\frac{1-\alpha }{1+\alpha }\right)+137.5\left(\frac{\alpha }{1+\alpha }\right)+71\left(\frac{\alpha }{1+\alpha }\right)=124$.

$\alpha =0.6815$.

Thus, percentage dissociation of $PC{l}_5$ is $68.15$%.