Question

2 litres of a mixture of nitrous and nitric oxides at STP have a mean molecular weight of 39.8. What volume
nitrogen measured at STP could be obtained when the mixture has been passed over red hot copper?
(A) 1.7 litres
(B) 1.9 litres
(C) 1.5 litres
(D) 1.85 litres​

Answer 1

Answer : The correct options is, (A) 1.7 liters

Explanation :

As we are given the volume of mixture of nitrous oxide and nitric oxide is 2L.

Let us assumed that the volume of nitric oxide be, 'x' liter and volume of nitrous oxide be, (2-x) liter.

Formula used :

$\bar M=\sum x_iM_i$

where,

$\bar M$ = mean molecular weight

$x_i$ = mole fraction

$M_i$ = molar mass

First we have to calculate the moles of $NO$ and $N_2O$

$\text{Moles of }NO=\frac{\text{Volume of }NO}{\text{Molar volume of }NO}=\frac{xL}{22.4L}$

$\text{Moles of }N_2O=\frac{\text{Volume of }N_2O}{\text{Molar volume of }N_2O}=\frac{(2-x)L}{22.4L}$

Now we have to calculate the mole fraction of $NO$ and $N_2O$

$\text{Mole fraction of }NO=\frac{\frac{xL}{22.4L}}{\frac{xL}{22.4L}+\frac{(2-x)L}{22.4L}}$

$\text{Mole fraction of }N_2O=\frac{\frac{(2-x)L}{22.4L}}{\frac{xL}{22.4L}+\frac{(2-x)L}{22.4L}}$

Now put all the given values in the above formula, we get:

$\bar M=x_{NO}\times M_{NO}+x_{N_2O}\times M_{N_2O}$

where,

$x_{NO}$ = mole fraction of NO

$x_{N_2O}$ = mole fraction of $N_2O$

$M_{NO}$ = molar mass of NO

$M_{N_2O}$ =molar mass of $N_2O$

Given : Molar mass of NO = 30 g/mole

Molar mass of $N_2O$ = 44 g/mole

Mean molecular weight = 39.8

$39.8=\frac{\frac{xL}{22.4L}}{\frac{xL}{22.4L}+\frac{(2-x)L}{22.4L}}\times 30+\frac{\frac{(2-x)L}{22.4L}}{\frac{xL}{22.4L}+\frac{(2-x)L}{22.4L}}\times 44$

By solving the terms, we get :

$x=0.61L$

The volume of NO = 0.61 L

The volume of $N_2O$ = 2 - x = 2 - 0.61 = 1.39 L

Now we have to calculate the volume of nitrogen at STP when the mixture has been passed over red hot copper.

The balanced chemical reactions are :

(1) $2NO+2Cu\rightarrow 2CuO+N_2$

As, $2\times 22.4L$ volume of NO react to give $1\times 22.4L$ volume of $N_2$

So, $0.61L$ volume of NO react to give $\frac{1\times 22.4L}{2\times 22.4L}\times 0.61L=0.305L$ volume of $N_2$

(2) $N_2O+Cu\rightarrow CuO+N_2$

As, 22.4 L volume of $N_2O$ react to give 22.4 L volume of $N_2$

So, 1.39 L volume of $N_2O$ react to give 1.39 L volume of $N_2$

The total volume of nitrogen = 0.305 L + 1.39 L = 1.697 L = 1.7 L

Therefore, the total volume of nitrogen is, 1.7 liters