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The vapour density of a mixture containing $NO_2$ and $N_2O_4$ is 38.3 at $27^{\large\circ}C$.Calculate the mole of $NO_2$ in 100g mixture

$(a)\;1.09\qquad(b)\;0.956\qquad(c)\;0.437\qquad(d)\;0.0045$

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Mol.wt of mixture of $NO_2$ and $N_2O_4$=38.3$\times 2=76.6$
Let 'a' g of $NO_2$ present in 100g mixture.
$\therefore$ Mole of $NO_2$+Mole of $N_2O_4$=Mole of mixture
$\large\frac{a}{46}+\frac{100-a}{92}=\frac{100}{76.6}$
$a=20.10g$
$\therefore $ Mole of $NO_2$ in mixture =$\large\frac{20.10}{46}$
$\Rightarrow 0.437$
Hence (c) is the correct answer.
answered Oct 30, 2013 by sreemathi.v
 

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