Question

A closed container of volume $0.02m^3$ contains a mixture of neon and argon gases at a temperature of $27^{\circ }C$ and pressure of $1\times 10^{5}N{m}^2$. The total mass of the mixture is $28g$. If the gram molecular weights of neon and argon are $20$ and $40$ , respectively, then find the mass (in $\rho$ ) of argon gas in the container. Assume gases to be ideal. (Given $R=8.314Jmo{l}^{-1}{K}^{-1}$ )

Answer 1

Answer: 24

Let $m$ gram be the mass of neon. Then the mass of argon is $(28-m)g$.
Total number of moles of the mixture,
$\mu =\frac{m}{20}+\frac{28-m}{40}=\frac{28+m}{40}...(i)$
Now, $\mu =\frac{PV}{RT}=\frac{1\times 10^5\times 0.02}{8.314\times 300}=0.8...(ii)$
From (i) and (ii),
$\frac{28+m}{40}=0.8$
$\Rightarrow 28+m=32$
$\Rightarrow m=4g$
or mass of argon $=(28-4)g=24g$