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# If, in the reaction $N_2O_4 \equiv 2NO_2$, x is that part of $N_2O_4$ which dissociates, then the number of molecules at equilibrium will be

$\begin{array}{1 1} (a)\;1 \\ (b)\;3 \\ (c) \;(1+x) \\ (d)\;(1+xy)^2 \end{array}$

$\qquad \qquad N_2O_4 \qquad \to \qquad 2 NO_2$
Begining : a moles $\qquad \qquad$ 0 moles
Moles lost /formed : ax lost $\qquad$ 2ax formed
At eqbm : a-ax $\qquad$ 2ax
Therefore , total $=a(1+x)$.
If initially 1 mole was taken , then no of moles $=1-x$
No of molecules $= N_a(1-x)$