A mixture of NO_2 and N_2O_4 has a vapour den | vedclass

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Question

(C) The molar mass of the mixture is calculated as $M_{mix} = 2 	imes 	ext{Vapour Density} = 2 	imes 38.3 = 76.6 \, g/mol$.
Let the number of mole

Vedclass · Accepted Answer

$33.48$

Vedclass · Answer

(C) The molar mass of the mixture is calculated as $M_{mix} = 2 	imes 	ext{Vapour Density} = 2 	imes 38.3 = 76.6 \, g/mol$.
Let the number of moles of $NO_2$ be $x$. Then the number of moles of $N_2O_4$ is $(100 - x)$.
The molar mass of $NO_2$ is $46 \, g/mol$ and $N_2O_4$ is $92 \, g/mol$.
The total mass of the mixture is given by the sum of the masses of the components:
$x(46) + (100 - x)(92) = 100 	imes 76.6$.
$46x + 9200 - 92x = 7660$.
$-46x = 7660 - 9200$.
$-46x = -1540$.
$x = \frac{1540}{46} \approx 33.48 \, moles$.

$A$ mixture of $NO_2$ and $N_2O_4$ has a vapour density of $38.3$ at $270,^oC$. Calculate the number of moles of $NO_2$ in $100$ moles of the mixture.

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