At `27^(@)C` and 1 atm pressure the volume of a 5.0 g mixture of He and Ar gases is `10dm^(3)`. Find the mass percent of the two gases in the gas mixture.

Explanation

let, the amount of He and Ar in the mixture be ag and bg respectively. Hence, a+b=5 . . . [1]
Now, `PV=(n_(1)+n_(2))RT or, 1xx10=[(a)/(4)+(b)/(40)]xx0.0821xx300`
`therefore 10a+b=16.24`
by solving [1] and [2] we get, `a=1.25 g and b=3.75g`
`therefore %He=(1.25)/(5)xx100=25 and %Ar=(3.75)/(5)xx100=75`