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 moles of nitrogen. What is the molar mass of unknown gas?

(Use $R=0.082$ L-atm/mol-k)

Answer 1

Here $V=4L$, $T=273K$, $R=0.082L-atm/mol-k$, $P=2.8atm$

So, total moles of solution

$n=\frac{PV}{RT}=\frac{2.8\times 4}{0.082\times 273}=0.5moles$

As moles of nitrogen gas $=0.4$

Hence, number of moles of unknown gas $=0.5-0.4=0.1moles$

Now from Grahm's law of dissusion

Rate of effusion of nitrogen $\left(r_N\right)=\frac{No.ofmoles}{Timetaken}=\frac{0.4}{10}=0.04mol/min$

Rate of effusion of unknown gas $\left(r_{unknown}\right)=\frac{0.1}{10}=0.01mol/min$

Further,

$\frac{r_N}{r_{unkown}}=\sqrt{\frac{m_{unknown}}{m_N}}$

$\frac{0.04}{0.01}=\sqrt{\frac{m_{unknown}}{m_N}}$

$4=\sqrt{\frac{m_{unknown}}{m_N}}$

Squaring both side we get,

$\frac{m_{unkown}}{m_n}=16$

$m_{unkown}=m_N\times 16$

Since molar mass of nitrogen gas $\left(N_2\right)=28$

$m_{unkown}=28\times 16=448gm/mol$