Question

Vapour density of $P{Cl}_5$ is 104.16 but when heated to ${230}^{o}C$ its vapour density is reduced to 62.The degree of dissociation of $P{Cl}_5$ at this temperature will be:

• A. 6.8%
• B. 68%
• C. 46%
• D. 64%

Answer 1

Answer: (B)

68%

The molar mass $PC{l}_3$ is $137.33$ g/mol
The molar mass $C{l}_2$ is $70.6$ g/mol

Initial molar mass $2\times$ initial vapour density
$=2\times 104.16=208.32g/mol$

Final molar mass $=2\times$ final vapour density
$=2\times 62=124g/mol$

Let out of 1 mole of $PC{l}_5$ present initially, $x$ mole dissociates to form $x$ mole of $PC{l}_3$ and x mole of $C{l}_2$

The molar mass of final mixture $=124=\frac{208.32(1-x)+137.33x+70.6x}{1-x+x+x}$

$124=\frac{208.32(1-x)+(137.33+70.6)x}{1+x}$

$124=\frac{208.32(1-x+x)}{1+x}$

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

$1+x=\frac{208.32}{124}=1.68$

$x=1.68-1=0.68$

The degree of dissociation of $PC{l}_5$ at ${230}^{\circ }C$ is $\frac{0.68}{1}=0.68$ or $68$ %.