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Q. $9.2\, g\, N_2O_4$ is heated in a $1L$ vessel till equilibrium state is established
$N_2O_{4(g)}\ce{<=>}2NO_{2(g)}$
In equilibrium state $50\% N_2O_4$ was dissociated, equilibrium constant will be (mol. wt. of $N_2O_4 = 92$)

UP CPMTUP CPMT 2008

Solution:

Moles of $N_{2}O_{4} =\frac{9.2}{92.0} =0.1$ mole
$\begin{matrix}&N_{2}O_{4}&\ce{<=>}&2NO_{2}(g)&&\\ \text{Initiailly }&0.1 {\text{moles}}&&0&\\ \text{At eqm.}&0.05&&0.05 \times 2 \end{matrix}$
$K=\frac{\left[NO_{2}\right]^{2}}{\left[N_{2}O_{4}\right]}$
$=\frac{0.1\times0.1}{0.05}$
$=0.2$