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$20\% N_2O_4$ molecules are dissociated in a sample of gas $27^{\large\circ}$ and 760 torr. Calculate the density of the equilibrium mixture.

$(a)\;2.116g/l\qquad(b)\;3.116g/l\qquad(c)\;1.11g/l\qquad(d)\;0.28g/l$

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$N_2O_4\Leftrightarrow 2NO_2$
Mole present initially in $N_2O_4$ = 1
Mole present initially in $2NO_2$ = 0
Mole at equilibrium $N_2O_4$ = (1-0.2)
Mole at equilibrium $2NO_2$ = 0.4
Molecular weight of the mixture = $\large\frac{0.8\times92+0.4\times46}{1.2}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= 76.66$
Now PV =$\large\frac{w}{m}RT$
$\large\frac{w}{V} = \large\frac{Pm}{RT} $
$= \large\frac{76.66\times1}{0.082\times300}$
= 3.116 g/l
Hence answer is (b)
answered Feb 14, 2014 by sharmaaparna1
 

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