Question

For $10$ minute each, at $0^o$C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of $4$ litre capacity. The resulting pressure is $2.8$ atm and the mixture contains $0.4$ mole of nitrogen. What is the molar mass of unknown gas? (Use $R=0.082$L-atm/mol-K).

Answer 1

Let the unknown gas be X.

Let $M_N$ and $M_X$ be the molar mass of nitrogen and the other gas respectively.

From ideal gas equation-

$PV=nRT$

Given that

$P=2.8\text{bar}$

$V=4L$

$T=0℃=273K$

$R=0.0821\text{L atm J}{K}^{-1}{mol}^{-1}$

$\therefore$ total no. of moles of solution $\left(n\right)=\frac{2.8\times 4}{0.0821\times 273}=0.5\text{moles}$

Given no. of moles of nitrogen gass $=0.4\text{moles}$

$\therefore$ No. of moles of gas X $=0.5-0.4=0.1\text{moles}$

Now from Graham's law of diffusion,

rate of effusion $=\frac{\text{no. of moles}}{\text{time taken}}$

Given that $t=10\text{minutes}$

Rate of effusion of nitrogen $\left(r_N\right)=\frac{0.4}{10}=0.04$

Rate of effusion of gas X $\left(r_X\right)=\frac{0.1}{10}=0.01$

Further, we know that

$\frac{r_N}{r_X}=\sqrt{\frac{M_X}{M_N}}$

As we know that molar mass of nitrogen gas is $28g$.

$\therefore \frac{0.04}{0.01}=\sqrt{\frac{M_X}{28}}$

$\Rightarrow \sqrt{M_X}=4\times \sqrt{28}$

Squaring bth sides, we have

$M_X=16\times 28=448gm/mol$

Hence the molar mass of unknown gas is $448gm/mol$.