Question

The vapour density of a mixture containing $N{O_2}$ and ${N_2}{O_4}$ is $38.3$ at $27^\circ C$. Calculate the moles of $N{O_2}$ in $100g$of the mixture.

Answer 1

Vapour density is equal to the density of a vapour compared to that of hydrogen. It is rather calculated as the ratio of mass of a certain volume of a substance divided by mass of the same volume of hydrogen.

Complete step by step answer: The mathematical equation of vapour density can be expressed as:
${V_d} = \dfrac{{{M_n}(g)}}{{{M_n}({H_2})}}$
where ${V_d}$ = vapour density,
${M_n}(g)$ = mass of n number of molecules of desired gas,
${M_n}({H_2})$ = mass of n number of molecules of hydrogen.
The molar mass of the mixture = $2 \times {V_d} = 2 \times 38.3 = 76.6.$
For $N{O_2}$ the molecular weight = molecular weight of $N$ + $2$ x molecular weight of $O$
= $14 + 2 \times 16 = 46$ .
For ${N_2}{O_4}$ the molecular weight = $2$ x molecular weight of $N$ + $4$ x molecular weight of $O$
= $2 \times 14 + 4 \times 16 = 92$ .
Let the total moles of mixture = $1$
Let the moles of $N{O_2}$ = $x$
The moles of ${N_2}{O_4}$ = $1 - x$
Moles of $N{O_2}$ + moles of ${N_2}{O_4}$ = moles of mixture
$46x + 92(1 - x) = 76.6$
$46x = 15.4$
$x = \dfrac{{15.4}}{{46}} = 0.335$
The percentage of $N{O_2}$ = 33.5
And the percentage of ${N_2}{O_4}$ = 66.5
Thus the number of moles of $N{O_2}$ in $100g$of the mixture = $0.335 \times \dfrac{{100}}{{76.6}}$
= $0.437$

Note: Density and vapour density should not be confused as density is the ratio of mass of a substance and the volume of the substance while vapour density is the ratio of weight obtained by the dividing volume of gas or vapour compared to the weight of an equal volume of hydrogen. The number of moles can be calculated by dividing the mass of substance and the substance's atomic or molecular weight. Similarly the mole fraction is determined by dividing the moles of one substance in a mixture by the total number of moles of all substances in the mixture.