Question

The relative density of a mixture of nitrogen and oxygen is 14.4 (H = 1) and the relative densities of nitrogen and oxygen are 14.0 and 16.0 (H =1)respectively. Calculate the composition of the mixture (i) by volume and (ii) by mass.

Answer 1

(i) Let 1 mL of the mixture contain x mL of $N_2$ and (1 - x) mL of $O_2$.
Mass of x mL of $N_2=140\times x=14xg$ (Mass $=d\times V$)
Mass of (1 - x) mL of $O_2=16.0\times (1-x)=(16.0-16x)g$
Total mass of mixture $=14x+16.0-16x$
So, $14x+16.0-16x=14.4\times 1=14.4$
or x = 0.8 , i.e., 80% by volume
Oxygen $=1-x=(1-0.8)=0.2$, i.e., 20% by volume.
(ii) Let 1 g of the mixture contain x g of $N_2$ and (1- x) g of oxygen.
Volume of x g of $N_2=\frac{x}{14.0}(V=\frac{mass}{density})$
Volume of (1-x) g of $O_2=\frac{(1-x)}{16.0}$
Total volume of the mixture $=\frac{1}{14.0}+\frac{(1-x)}{16.0}$
So, $=\frac{1}{14.0}+\frac{(1-x)}{16.0}=\frac{1}{14.4}$
or x=0.7778, i.e., 77.78% by mass
Oxygen $=(1-x)=(1-0.7778)=0.2222$
i.e., $22.22\%$ by mass.