Question

A synthetic mixture of nitrogen and Argon has a density of $1.4g{L}^{-1}$ at $0^{o}C$. Find out the volume percentage of nitrogen in the mixture?

• A. $85$
• B. $96$
• C. $72$
• D. $60$

Answer 1

The correct option is C $72$

Density of $Ar$ at $0^{\circ }C=1.7818$

Density of $N_2$ at $0^{\circ }C=1.2504$

$\Rightarrow 1.4=X_{N_2}\left(1.2504\right)+(1-X_{N_2})1.7818$

$\therefore X_{N_2}=\frac{0.3818}{0.5314}=72\%$

$\because V=\frac{nRT}{P}\Rightarrow Volume\%=\%X_{N_2}$

$\therefore$ Volume of $N_2=72\%$