$PCl_{5}$ dissociates as $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. $5 \, \text{moles}$ of $PCl_{5}$ are placed in a $200 \, L$ vessel which contains $2 \, \text{moles}$ of $N_{2}$ and is maintained at $600 \, K$. The equilibrium pressure is $2.46 \, atm$. The equilibrium constant $K_{p}$ for the dissociation of $PCl_{5}$ is $...... \times 10^{-3}$. (nearest integer) (Given: $R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$: Assume ideal gas behaviour)

  • A
    $2312$
  • B
    $954$
  • C
    $1107$
  • D
    $1451$

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At $T$ $(K)$,$K_c$ for the dissociation of $PCl_5$ is $2 \times 10^{-2} \ mol \ L^{-1}$. The number of moles of $PCl_5$ that must be taken in a $1.0 \ L$ flask at the same temperature to get $0.2 \ mol$ of chlorine at equilibrium is:

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