Question

At $0^{o}C$ the density of certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of oxide?

Answer 1

From ideal gas equation,

$PV=nRT$

$\because n=\frac{\text{Weight}\left(W\right)}{\text{Mol. wt.}\left(M\right)}$

$\Rightarrow PV=\left(\frac{W}{M}\right)RT$

$\Rightarrow \frac{W}{V}\frac{PM}{RT}$

$\because \text{Density}\left(\rho \right)=\frac{W}{V}$

$\therefore \rho =\frac{PM}{RT}.....\left(1\right)$

For oxide:-

Let $M$ be the molecular mass of oxide.

$P=2bar\left(\text{Given}\right)$

Therefore,

${\rho }_1=\frac{2M}{RT}$

For dinitrogen:-

$P=5bar$

Molecular mass of $N_2=28g$

Therefore,

${\rho }_2=\frac{5\times 28}{RT}=\frac{140}{RT}$

$\because {\rho }_1={\rho }_2\left(\text{Given}\right)$

$\therefore \frac{2M}{RT}=\frac{140}{RT}$

$\Rightarrow M=\frac{140}{2}=70g$

Hence the olecular mass of oxide is $70g/mol$.