A closed container of volume `0.02m^(3)` contains a mixture of neon and argon gases at a temperature `27^(@)C` and pressure `1 xx 10^(5) Nm ^(-1)` The total mass is 28 and the molar mass of and argon are `20` and `40` respectively find the masses of individual gases in the container assuming then to be ideal .

Options

A. 24 g
B. 25 g
C. 26 g
D. 27 g

Answer

A

Explanation

Let in the given container mass of neon be m and mass of argon be (28-m) g, so that
`n_("Ne")=(m)/(20)and n_(A)=(28-m)/(40)`
`n=n_("Ne")+n_(A)=(m)/(20)+((28-m))/(40)=(28+m)/(40)` …. (i)
and using ideal gas equation for the mixture, we have
`n=(pV)/(RT)=(1xx10^(5)xx0.02)/(8.314xx300)=0.8` (ii)
Comparing Eqs .(i) and (ii) , we get
`(28+m)/(40)=0.8rArrm=4g`
`therefore " " m_(Ne)=4gand m_(A)=28-4=24g`