Question

The vapour density of $PC{l}_5$ is 104.16 but when heated to ${230}^{o}C$, its vapour density is reduced to 62. The degree of dissociation of $PC{l}_5$, at ${230}^{o}C$, is:

• A. $0.34$
• B. $0.68$
• C. $0.80$
• D. None of these

Answer 1

The correct option is B $0.68$

The vapour density of $PC{l}_5$ is 104.16 but when heated to ${230}^{o}C$, its vapour density is reduced to 62. The degree of dissociation of $PC{l}_5$

, at ${230}^{o}C$ is $0.68$
The molecular weight is twice the vapour density.
When the vapor density is 104.16, the molecular weight is $2\times 104.16=208.32$ g/mol.
When the vapor density is 62, the molecular weight is $2\times 64=124$ g/mol
Let x be the degree of dissociation of $PC{l}_5$

	$PC{l}_5$	$PC{l}_3$	$C{l}_2$
Initial moles	1	0	0
Equilibrium moles	$1-x$	$x$	$x$

$\frac{208.32(1-x)+137.33x+71x}{1-x+x+x}=124$
$\frac{208.32}{1+x}=124$
$208.32=124+124x$
$84.32=124x$
$x=0.68$
Hence, the degree of dissociation of $PC{l}_5$ is $0.68$.