Question

Vapor density of mixture of $N{O_2}\& {N_2}{O_4}$ is 34.5, then percentage abundance of $N{O_2}$ in the mixture is:
A) $50\%$
B) $25\%$
C) $40\%$
D) $60\%$

Answer 1

The relationship between Vapor density and Molar Mass of the Mixture can be given as:
$V.{D_{Mixture}} = \dfrac{{{M_{mixture}}}}{{{M_{{H_2}}}}}$ ; Where ${M_{mixture}}$ is the molar mass of the Mixture of Gases and ${M_{{H_2}}}$ is the Molar mass of Hydrogen Molecule. Simplifying the above equation $V.{D_{Mixture}} = \dfrac{{{M_{mixture}}}}{2}$

Complete answer: We are given a mixture of two gases $N{O_2}\& {N_2}{O_4}$ . Consider the mole fraction of both the gases to be ${\chi _{N{O_2}}}\& {\chi _{{N_2}{O_4}}}$ respectively.
The vapor density of the mixture of gases is given as 34.5, finding the molar mass of the mixture from the formula given above: ${M_{mixture}} = 2 \times V.{D_{mixture}} = 2 \times 34.5 = 69g$
Mole fraction of $N{O_2} = {\chi _{N{O_2}}}$
Mole fraction of ${N_2}{O_4} = {\chi _{{N_2}{O_4}}}$
The total of mole fraction will always be unity. ${\chi _{N{O_2}}} + {\chi _{{N_2}{O_4}}} = 1 \to {\chi _{{N_2}{O_4}}} = 1 - {\chi _{N{O_2}}}$
The molar mass of any mixture can be given as: ${M_{mix}} = {\chi _1}{M_1} + {\chi _2}{M_2}$ ( where ${M_1}\& {M_2}$ are the molar masses of the constituent gases)
Using the above formula, the Molar mass of the mixture of $N{O_2}\& {N_2}{O_4}$ can be given as:
${M_{mixture}} = {M_{N{O_2}}} \times {\chi _{N{O_2}}} + {M_{{N_2}{O_4}}} \times {\chi _{{N_2}{O_4}}}$
${M_{mixture}} = {M_{N{O_2}}} \times {\chi _{N{O_2}}} + {M_{{N_2}{O_4}}} \times (1 - {\chi _{N{O_2}}})$
The Molar Mass of $N{O_2}\& {N_2}{O_4}$ are 46 and 92 respectively. Substituting the values, we get:
$69 = 46 \times {\chi _{N{O_2}}} + 92 \times (1 - {\chi _{N{O_2}}})$
$69 - 92 = 46{\chi _{N{O_2}}} - 92{\chi _{N{O_2}}}$
${\chi _{N{O_2}}} = \dfrac{1}{2}$
Now, we have to find the percentage abundance from mole fraction.
The percentage abundance of $N{O_2} = \dfrac{1}{2} \times 100 = 50\%$
The correct answer is Option (A).

Note:
There is also an alternate way to solve the given problem. Consider the molar masses of $N{O_2}\& {N_2}{O_4}$ to be 46 and 92 respectively. The mass of the mixture can be found by the same formula above that gives it to be 69g.
The total mass of the Mixture can also be found out by a different formula, if we consider the total no. of moles to be 100. The no. of moles of $N{O_2}\& {N_2}{O_4}$ will be $x\& (100 - x)$ respectively. The total mass of the mixture can be given as: ${M_{mixture}} = \dfrac{{{n_{N{O_2}}} \times {M_{N{O_2}}} + {n_{{N_2}{O_4}}} \times {M_{{N_2}{O_4}}}}}{{Total{\text{ }}moles}}$
$69 = \dfrac{{x \times 46 + (100 - x) \times 92}}{{100}}$ $\dfrac{{{n_{N{O_2}}}}}{{Total{\text{ }}moles}} \times 100 \to \dfrac{{50}}{{100}} \times 100 = 50\%$