Question

A mixture of \[N{O_2}\] and \[{N_2}{O_4}\] has a vapour density of 38.3 at 300K. What is the number of moles of \[N{O_2}\] in 100g of the mixture?
A.0.043
B.4.4
C.3.4
D.0.437

Answer 1

Hint: To solve this question, we must first find the molar mass of the mixture. Then we must find the ratios of the constituent gases present in the mixture on the basis of the number of moles of each gas. Then, in the end, we can find the ratio of the number of moles of \[N{O_2}\] present in 1 mol of the mixture and multiply it by 100 g to get the final answer.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
When we have to calculate the vapour density of a mixture of gases, we can use the following relation:
Vapour density \[ = \dfrac{1}{2}\] (molar mass of the mixture)
Hence, in the given question, the molar mass of the mixture of \[N{O_2}\] and \[{N_2}{O_4}\] can be calculated as:
Molar mass = 2 (Vapour density) = 2 (38.3) = 76.6
Ow, let the total number of moles of gases present in the mixture be 1. Then, let us consider the number of moles of \[N{O_2}\] to be ‘n’. hence, the number of moles of \[{N_2}{O_4}\] is \[\left( {1-n} \right)\] . Since the molar mass of the mixture is given to be 76.6 g/mol, and the molar masses of \[{N_2}{O_4}\] and \[N{O_2}\] are 46 and 92 respectively, we can say that:
 \[\begin{array}{*{20}{l}}
  {n + \left( {1 - n} \right) = 1} \\
  {n\left( {46} \right) + \left( {1 - n} \right)\left( {92} \right) = 76.6} \\
  {46n = 15.4} \\
  {n = 0.335}
\end{array}\]
Hence, the number of moles of \[N{O_2}\] present in 100g of the mixture can be given as:
 \[X = \dfrac{{no.\,of\,moles\,of\,N{O_2}\,present\,in\,1\,mole\,of\,mixture}}{{molar\,mass\,of\,mixture}} \times 100 = \dfrac{{0.335}}{{76.6}} \times 100\] = 0.437

Hence, Option D is the correct option

Note: There is another way in which we can find the vapour density of a gas. When we find the ratio of the mass of a certain volume of gas to the mass of an equal volume of hydrogen under the same conditions of temperature and pressure, we get a relation known as vapour density. This quantity can be mathematically represented as:
Vapour density \[ = \dfrac{{weight\,of\,a\,certain\,volume\,of\,gas}}{{weight\,of\,an\,equal\,volume\,of\,{H_2}}}\]