In an evaculated closed isolated chamber at $250^{\circ} Co . o 2$ mole ${PCl}_{5}$ and $0.1 {mole}_{2}$ are mixed. $({PCl}_{5} \Leftrightarrow {PCl}_{3} + {Cl}_{2})$ At equilibrium, density of mixture was $2.48 g / L$ and pressure was 1atm. The number of total moles at equilibrium will be approximately :

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0.012

0.106

0.022

0.045

answer is C.

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Detailed Solution

$\begin{array}{cccc} {PCl}_{5} & ⇌ {PCl}_{3} + & {Cl}_{2} \\ 0.02 & - & 0.1 \\ 0.02 - x & x & 0.1 + x \end{array}$

$\begin{array}{l} P \cdot M = d \cdot R \cdot T \\ M = 2.48 \times 0.0821 \times 523 \\ = 106.5 \\ (0.1 \times 71) + (0.02 \times 208.5) \\ = 106.5 \times n \\ n = 0.106 \end{array}$

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