Question

Vapour density of $PC{l_5}$ is $104.16$ but when heated to $230^\circ C$ its vapour density is reduced to $62$. The degree of dissociation of $PC{l_5}$ at this temperature will be:
A. $6.8\% $
B. $68\% $
C. $46\% $
D. $64\% $

Answer 1

Hint: Vapour density is defined in relation to hydrogen. When the mass of a certain volume of any substance is divided by the mass of a volume of hydrogen. It can be represented as follows:
Vapour density $ = $ mass of n molecules of gas/ mass of n molecules of hydrogen.
If we talk about the degree of dissociation, it can be defined as the mole of the fraction undergoing dissociation at a given time or temperature. It is represented by $\alpha $
$\alpha = \dfrac{{number \;of \;moles\; dissociated}}{{number \;of \;moles\; present\; initially}}$

Complete step by step answer:
In the question, the compound given is $PC{l_5}$. The molar mass is also related to the vapour density. The relation between the molar mass and the vapour density can be written as:
Molar mass $ = 2 \times vapour. density$
When $PC{l_5}$ is dissociated into $PC{l_3}$. The reaction is
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
The molar mass of $PC{l_3}$ is $137.33g/mol$
The molar mass of $C{l_2}$ is $70.6g/mol$
Now, according to the mentioned relation, we can calculate the initial and final molar mass with the help of given values of vapour density.
Initial molar mass $ = 2 \times 104.16 = 208.32g/mol$
Final molar mass $ = 2 \times 62 = 124g/mol$
If we look at the reaction then we can assume that 1 mole of $PC{l_5}$ dissociates into y mole of $PC{l_3}$and y mole of $C{l_2}$
So, the final molar mass can be written as:
Molar mass $ = \dfrac{{208.32(1 - y) + 137.33y + 70.6y}}{{1 - y + y + y}}$
Molar mass is $124g/mol$
After substituting the values we have
$124 = \dfrac{{208.32(1 - y) + 137.33y + 70.6y}}{{1 + y}}$
$ \Rightarrow 124 = \dfrac{{208.32(1 - y + y)}}{{1 + y}}$
$ \Rightarrow 124 = \dfrac{{208.32}}{{1 + y}}$
$ \Rightarrow y = 0.68$
So, we can conclude that the degree of dissociation at $230^\circ C$ will be $0.68\;or\;68\% $

So, the correct answer is Option B.

Note: There is an alternate method to find the degree of dissociation.
Alternate method:
Let us write down the reaction i.e.
 $PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$

	$PC{l_5}$	$PC{l_3}$	$C{l_2}$
Initial	C	0	0
At equilibrium	$C(1 - \alpha )$	$C\alpha $	$C\alpha $

Here, $\alpha $ represents the degree of dissociation.
Now, we will calculate the total number of moles
Total moles $ = C(1 - \alpha ) + C\alpha + C\alpha $
Total moles $ = C(1 + \alpha )$
As we know that at the equilibrium vapour density is inversely related to the number of moles.
Thus, we can write it as $\dfrac{{Total\; Mole\; At\; Equilibrium}}{{Initial. Moles}} = \dfrac{{Initial \;Vapour \;Density}}{{Final\; Vapour\; Density}}$
After substituting the values from the table and the question
$\dfrac{{C(1 + \alpha )}}{C} = \dfrac{{104.16}}{{62}}$
$ \Rightarrow 1 + \alpha = \dfrac{{104.16}}{{62}}$
$ \Rightarrow \alpha = \dfrac{{104.16}}{{62}} - 1$
$ \Rightarrow \alpha = 0.68\;or\;68\% $
Thus, the degree of dissociation is $0.68\;Or\;68\% $