Answer

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**Hint:**Degree of dissociation is defined as fraction of the mole of reactant that got dissociated. It is represented by \[\alpha \].

\[\alpha \text{ = }\dfrac{\text{amount of reactant dissociated}}{\text{amount of reactant present initially}}\]

**Complete step by step answer:**In order to find density, we will first find the number of moles in the reaction. We know that \[\alpha = 0.4\].

We will calculate both, the number of moles present initially and the number of moles present at equilibrium.

\[\begin{matrix}

{} \\

\text{Number of moles initially -} \\

\text{Number of moles at equilirbium - } \\

\end{matrix}\begin{matrix}

PC{{l}_{5}} \\

1 \\

(1-0.4) \\

\end{matrix}\begin{matrix}\leftrightarrows

PC{{l}_{3}} \\

1 \\

0.4 \\

\end{matrix}+\begin{matrix}

C{{l}_{2}} \\

1 \\

0.4 \\

\end{matrix}\]

\[\begin{align}

& \text{Total = 0}\text{.6 + 0}\text{.4 + 0}\text{.4} \\

& \text{ = 1}\text{.4} \\

& \text{ n = 1}\text{.4} \\

\end{align}\]

We know that the ideal gas equation is \[\text{PV = nRT}\]

But we have to find density, for which we need volume. Therefore, \[\text{V = }\dfrac{\text{nRT}}{\text{P}}\]

We have been given that \[\text{P = 1 atm}\] and

So, we take \[\text{R = 0}\text{.0821 L atm }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}\].

Substituting these values in the given equation we get,

\[\begin{align}

& \text{V = }\dfrac{\text{nRT}}{\text{P}} \\

&\implies \text{ }\dfrac{1.4\text{ }\times 0.0821 \times 400\text{ }}{1\text{ atm}} \\

& \text{ V = 45}\text{.976 L} \\

\end{align}\]

Now, we will calculate the mass of \[\text{PC}{{\text{l}}_{5}}\]. Mass always remains constant.

We have, \[\text{P = 31}\text{.0 Cl = 35}\text{.5}\]

\[\begin{align}

& \text{total = 31}\text{.0 + 5}\times \text{35}\text{.5} \\

&\implies \text{ 31}\text{.0 + 177}\text{.5} \\

&\implies \text{ 208}\text{.5 g} \\

\end{align}\]

Now, we know the formula for density.

\[\begin{align}

& \text{Density = }\dfrac{\text{mass}}{\text{volume}} \\

&\implies \text{ }\dfrac{208.9}{45.976} \\

&\implies \text{ 4}\text{.54 g }{{\text{L}}^{-1}}\text{ }\approx \text{ 4}\text{.53 g }{{\text{L}}^{-1}} \\

\end{align}\]

**Hence, option A is correct.**

**Additional information:**Reactants and products coexist in equilibrium so that reactant conversion to product is always less than 100%. Equilibrium reactions may involve the decomposition of a covalent reactant or ionization of ionic compounds into their ions in polar solvents.

**Note:**Since, we were given that \[\alpha \text{ = 0}\text{.4}\], to find the total number of moles, we subtracted it from 1. Also, remember that mass remains constant.

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