Question

The vapor density of a mixture containing $NO_2$ and $N_2O_4$ is $27.6$.The mole fraction of $NO_2$ in the mixture will be
A.$0.2$
B.$0.4$
C.$0.6$
D.$0.8$

Answer 1

Vapor density is defined as the weight of a volume of pure vapor or gas compared to an equal volume of dry air at the same temperature. Vapor density is obtained by dividing the molecular weight of the vapor by the average molecular weight of dry air, so it is unitless.

Formula used: Ideal gas law equation is given by ${\text{PV = nRT}}$
where ${\text{P }}$- Pressure
${\text{V}}$- Volume
${\text{T}}$- Temperature
${\text{n}}$-Number of moles
${\text{R}}$-Universal gas constant
Hence the vapor density can be calculated using the equation
Vapor density = $\dfrac{{\left( {\dfrac{{{\text{PM}}}}{{{\text{RT}}}}} \right)\,{\text{gas}}}}{{\left( {\dfrac{{{\text{PM}}}}{{{\text{RT}}}}} \right)\,{\text{reference}}\,{\text{gas}}}}$

Complete step by step solution:
Vapor density can be calculated using ideal gas law
${\text{PV = nRT}}$
Relative density for a gas is given by,
Relative density= $\dfrac{{{\text{Density}}\,{\text{of}}\,{\text{gas}}}}{{{\text{Density}}\,{\text{of}}\,{\text{reference}}\,{\text{gas}}}}$ ( at constant temperature and pressure)
Density of gas = $\dfrac{{{\text{given}}\,\,{\text{weight}}\,{\text{of}}\,{\text{gas}}}}{{{\text{volume}}\,{\text{of}}\,{\text{gas}}}}$ = $\dfrac{{\text{m}}}{{\text{V}}}$
From ideal gas equation we get the volume of gas
That is ${\text{V}}\,{\text{ = }}\,\dfrac{{{\text{nRT}}}}{{\text{P}}}$.
By applying the formula of volume , we get
Vapor density = $\dfrac{{{\text{mP}}}}{{{\text{nRT}}}}$ , where ${\text{Molecular}}\,{\text{weight}}\,{\text{M = }}\,\dfrac{{\text{m}}}{{\text{n}}}$
So that we can rearrange the equation as
Density = $\dfrac{{{\text{PM}}}}{{{\text{RT}}}}$
Hence Relative vapor density can be written as
Vapor density = $\dfrac{{\left( {\dfrac{{{\text{PM}}}}{{{\text{RT}}}}} \right)\,{\text{gas}}}}{{\left( {\dfrac{{{\text{PM}}}}{{{\text{RT}}}}} \right)\,{\text{reference}}\,{\text{gas}}}}$
At constant temperature and pressure we get
Vapor density = $\dfrac{{{\text{M}}\,{\text{gas}}}}{{{\text{M}}\,{\text{reference}}\,{\text{gas}}}}$
We are using $H_2$ as the reference gas so the molecular weight is 2.
So that the equation becomes, vapor density= $\dfrac{{{\text{M}}\,\,{\text{gas}}}}{{\text{2}}}$ = $\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{n}}\,{\text{molecules}}\,{\text{of}}\,{\text{gas}}}}{{\text{2}}}$
Here we have to calculate the mole fraction of $NO_2$ In a mixture containing $NO_2$ and $N_2O_4$ .where the vapor density is given by $27.6$
Vapor density = $\dfrac{{{\text{Molecular}}\,{\text{weight}}\,}}{{\text{2}}}$
Vapor density = $27.6$ , so Molecular weight = ${\text{vapor}}\, \text{density} \times 2$
Molecular weight= $27.6 \times 2\, = \,55.2$
Now consider % of $NO_2$ is ${\text{x}}$ and that of $N_2O_4$ is $\left( {{\text{1 - x}}} \right)$ . because sum of mole fractions of all components is equal to one. That is if a solution contains two components having mole fractions ${x_1}\,$and ${x_2}$ then ${x_1} + {x_2} = 1$.
So here Molecular weight of mixture containing the $NO_2$ and $N_2O_4$ can be
${\text{M = }}\left( {{\text{x}}\,\times \,{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{N}}{{\text{O}}_{\text{2}}}} \right){\text{ + }}\left( {\left( {{\text{1 - x}}} \right) \times \,{\text{Molar mass}}\,{\text{of}}\,{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}} \right)$= $55.2$
Molecular mass of $NO_2$ = $46$
Molecular mass of $N_2O_4$ = $92$
$\,\,\,{\text{x}}\left( {{\text{46}}} \right)\,{\text{ + }}\left( {{\text{1 - x}}} \right)\,{\text{92}}$ = 55.2
By solving above equation we get, ${\text{x}}$= $0.8$

The mole fraction of $NO_2$ in the mixture of $NO_2$ and $N_2O_4$ is $0.8$

Note: The value weight of a gas or vapor compared to air , which has an arbitrary value of one . If a gas has a vapor density of less than one it will generally rise in air. If the vapor density is greater than one gas will generally sink in air.