9.2 g of N_2O_{4(g)} is taken in a 1 L clos | vedclass

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(A) Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$.Given that $50\%$ of $N_2O_4$ dissociates,the moles reacted $= 0.1 	ime

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$0.2$

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(A) Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$.Given that $50\%$ of $N_2O_4$ dissociates,the moles reacted $= 0.1 	imes 0.5 = 0.05 \ mol$.$\begin{array}{lccc} & N_2O_{4(g)} & ightleftharpoons & 2NO_{2(g)} \ 	ext{Initial moles} & 0.1 & & 0 \ 	ext{Change} & -0.05 & & +2(0.05) \ 	ext{Equilibrium moles} & 0.05 & & 0.1 \end{array}$Since the volume is $1 \ L$,the concentrations are $[N_2O_4] = 0.05 \ mol \ L^{-1}$ and $[NO_2] = 0.1 \ mol \ L^{-1}$.$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2 \ mol \ L^{-1}$.

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