Question

A volume of 100 ml of air containing only oxygen and nitrogen is taken in a jar over water. $\text{NO}$ is slowly passed till no more brown fumes appear in the gas jar. It is found that 42 ml of $\text{NO}$ is required. The percentage of nitrogen in the air would be ______________.
A. $42%$
B. $79%$
C. $21%$
D. $39.5%$

Answer 1

Hint: Connect the stoichiometric coefficients of gases with their volume of the gases. The brown fumes of the gas mean there is a reaction between the gases to form new gas. The gas formed will be nitrogen dioxide $\left( \text{N}{{\text{O}}_{2}} \right)$ because of brown fumes. Find the volume of ${{\text{O}}_{2}}$ and then, the composition of gases will give the volume of ${{\text{N}}_{2}}$ and then, percentage of the gas.

Complete step by step answer:
Write the reaction between the reaction of oxygen gas and $\text{NO}$, as $\text{NO}$does not react with nitrogen gas to form any compound. The reaction is $\text{NO}+\dfrac{1}{2}{{\text{O}}_{2}}\to \text{N}{{\text{O}}_{2}}$. The compound formed will be nitrogen dioxide.

Let us solve this numerical question step by step:
Step (1)- Now, relate the volume of gases required with their respective stoichiometric coefficients. The coefficient of gas $\text{NO}$ is 1 in the reaction.
The coefficient of gas ${{\text{O}}_{2}}$ is ${1}/{2}\;$ in the reaction.
It can be written as $1\text{ NO=}\dfrac{1}{2}{{\text{O}}_{2}}$.
Step (2)- We need to relate the volume of $\text{NO}$and volume of oxygen gas.
The volume of $\text{NO}$ required is 42 ml (given).
The volume of ${{\text{O}}_{2}}$ required is $\dfrac{1}{2}\times 42$ or 21 ml.
Step (3)- The volume of ${{\text{N}}_{2}}$ will be $\left( 100-\text{volume of }{{\text{O}}_{2}} \right)$ as the total volume of the mixture is 100 ml.
The volume of ${{\text{O}}_{2}}$ is 21 ml. The volume of ${{\text{N}}_{2}}$ will be 100-21 or 79 ml.
Step (4)- The percentage of nitrogen gas will be $\dfrac{\text{volume of }{{\text{N}}_{2}}}{\text{total volume}}\times 100$ or $\dfrac{79}{100}\times 100$ which equals $79%$.

The correct option is option ‘b’.

Note: The moles of the gases can be related with the volume by the law, which is Avogadro law, that equal volumes of the gas contain the equal number of molecules at same temperature and pressure.