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The molecular mass of $PC{{l}_{5}}$ is $208.3$. At ${{200}^{o}}C$, the mass of partial dissociated vapours is $62$ times that of the mass of hydrogen. Calculate degree of dissociation of $PC{{l}_{5}}$.

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Last updated date: 23rd Jul 2024
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Answer
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Hint: The fraction of the compound which is dissociated into its respective ions or molecules is known as the degree of dissociation of that compound. In simple words, the degree of dissociation is the extent to which the dissociation of a compound occurs. It is represented by the symbol $\alpha $.

Complete answer:
To calculate the degree of dissociation for phosphorus pentachloride, first we need to find the relation between vapor density and the degree of dissociation as follows:
For the reaction, $A\rightleftharpoons nB$
Assuming initial moles of reactant is $x$. So, the number of moles of reactants at equilibrium will be $x(1-\alpha )$and the number of moles of the product will be $nx\alpha $. At equilibrium, the vapor density of the system is inversely proportional to the number of moles of gaseous species. Therefore,
$\dfrac{\text{initial vapor density}}{\text{vapor density at equilibrium}}=\dfrac{\text{number of moles at equilibrium}}{\text{initial number of moles}}$
Substituting values according to the given conditions:
$\Rightarrow \dfrac{D}{\text{d}}=\dfrac{x(1-\alpha )+nx\alpha }{x}$
$\Rightarrow \dfrac{D}{\text{d}}=\dfrac{x[1+\alpha (n-1)]}{x}$
\[\therefore \alpha =\dfrac{D-d}{(n-1)d}\,\,-(i)\]
Where, D is the initial vapor density, d is the vapor density at equilibrium, n is the number of products formed after dissociation.
Now, the dissociation of $PC{{l}_{5}}$ takes place as follows:
$PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$
As per given conditions, initial vapor density of $PC{{l}_{5}}=\dfrac{\text{molecular mass of }PC{{l}_{5}}}{2}$
$\Rightarrow D=\dfrac{208.3}{2}$
$\Rightarrow D=104.15$
Vapor density of $PC{{l}_{5}}$ at equilibrium $=62$
The number of products formed after reaction i.e., $n=2$
Substituting values in equation $(i)$:
\[\alpha =\dfrac{104.15-62}{(2-1)62}\]
$\therefore \alpha =0.6798$
In percent, the value of degree of dissociation $=0.6798\times 100\Rightarrow 67.9\%$
Hence the percentage of degree of dissociation of $PC{{l}_{5}}$ is $67.9\%$.

Note:
According to Avogadro’s law, volume is directly proportional to the number of moles of gas in the system. But we also know that volume is inversely proportional to the density of a molecule. Hence, on the basis of these two statements, we can say that vapor density inversely varies with the number of moles of the species present.