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Question: Vapour density of \(PC{l_5}\) is \(104.16\) but when heated to \(230^\circ C\) its vapour density is...

Vapour density of PCl5PC{l_5} is 104.16104.16 but when heated to 230∘C230^\circ C its vapour density is reduced to 6262. The degree of dissociation of PCl5PC{l_5} at this temperature will be:
A. 6.8%6.8\%
B. 68%68\%
C. 46%46\%
D. 64%64\%

Explanation

Solution

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 α\alpha
α=number  of  moles  dissociatednumber  of  moles  present  initially\alpha = \dfrac{{number \;of \;moles\; dissociated}}{{number \;of \;moles\; present\; initially}}

Complete step by step answer:
In the question, the compound given is PCl5PC{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×vapour.density = 2 \times vapour. density
When PCl5PC{l_5} is dissociated into PCl3PC{l_3}. The reaction is
PCl5⇌PCl3+Cl2PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}
The molar mass of PCl3PC{l_3} is 137.33g/mol137.33g/mol
The molar mass of Cl2C{l_2} is 70.6g/mol70.6g/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 =2×104.16=208.32g/mol = 2 \times 104.16 = 208.32g/mol
Final molar mass =2×62=124g/mol = 2 \times 62 = 124g/mol
If we look at the reaction then we can assume that 1 mole of PCl5PC{l_5} dissociates into y mole of PCl3PC{l_3}and y mole of Cl2C{l_2}
So, the final molar mass can be written as:
Molar mass =208.32(1−y)+137.33y+70.6y1−y+y+y = \dfrac{{208.32(1 - y) + 137.33y + 70.6y}}{{1 - y + y + y}}
Molar mass is 124g/mol124g/mol
After substituting the values we have
124=208.32(1−y)+137.33y+70.6y1+y124 = \dfrac{{208.32(1 - y) + 137.33y + 70.6y}}{{1 + y}}
⇒124=208.32(1−y+y)1+y\Rightarrow 124 = \dfrac{{208.32(1 - y + y)}}{{1 + y}}
⇒124=208.321+y\Rightarrow 124 = \dfrac{{208.32}}{{1 + y}}
⇒y=0.68\Rightarrow y = 0.68
So, we can conclude that the degree of dissociation at 230∘C230^\circ C will be 0.68  or  68%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.
PCl5⇌PCl3+Cl2PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}

| PCl5PC{l_5}| PCl3PC{l_3}| Cl2C{l_2}
---|---|---|---
Initial| C| 0| 0
At equilibrium| C(1−α)C(1 - \alpha )| CαC\alpha | CαC\alpha

Here, α\alpha represents the degree of dissociation.
Now, we will calculate the total number of moles
Total moles =C(1−α)+Cα+Cα= C(1 - \alpha ) + C\alpha + C\alpha
Total moles =C(1+α) = 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 Total  Mole  At  EquilibriumInitial.Moles=Initial  Vapour  DensityFinal  Vapour  Density\dfrac{{Total\; Mole\; At\; Equilibrium}}{{Initial. Moles}} = \dfrac{{Initial \;Vapour \;Density}}{{Final\; Vapour\; Density}}
After substituting the values from the table and the question
C(1+α)C=104.1662\dfrac{{C(1 + \alpha )}}{C} = \dfrac{{104.16}}{{62}}
⇒1+α=104.1662\Rightarrow 1 + \alpha = \dfrac{{104.16}}{{62}}
⇒α=104.1662−1\Rightarrow \alpha = \dfrac{{104.16}}{{62}} - 1
⇒α=0.68  or  68%\Rightarrow \alpha = 0.68\;or\;68\%
Thus, the degree of dissociation is 0.68  Or  68%0.68\;Or\;68\%