# Question be045

May 10, 2017

The mole fraction of ${\text{NO}}_{2}$ in the mixture is 0.791.

#### Explanation:

Vapour density is the density of a gas relative to that of hydrogen at the same temperature and pressure.

$\text{Vapour density" = "density of mixture"/("density of H"_2) = "mass of mixture"/("mass of H"_2) = "molar mass of mixture"/("molar mass of H"_2) = "Molar mass of mixture"/"2.016 g/mol} = 27.6$

$\text{Molar mass of mixture" = "27.6 × 2.016 g/mol" = "55.64 g/mol}$

Let $x$ = the mole fraction of ${\text{NO}}_{2}$.

Then $1 - x$ = the mole fraction of ${\text{N"_2"O}}_{4}$.

x × 46.01 color(red)(cancel(color(black)("g/mol"))) + (1-x) × 92.01 color(red)(cancel(color(black)("g/mol"))) = 55.64color(red)(cancel(color(black)( "g/mol")))#

$46.01 x + 92.01 \left(1 - x\right) = 55.64$

$46.01 x + 92.01 - 92.01 x = 55.64$

$46.00 x = 36.37$

$x = \frac{36.37}{46.00} = 0.791$

The mole fraction of ${\text{NO}}_{2}$ is 0.791.