The average molar mass of a mixture of methane `(CH_(4))` and ethene `(C_(2)H_(4))` present in the ratio of `a:b` is found to be `20.0g mol^(-1)`. If the ratio were reversed, what would be the molar mass of the mixture ?

Explanation

Molar mass of `CH_(4)="16 g mol"^(-1)`
`"Molar mass of "C_(2)H_(4)="28 g mol"^(-1)`
When they are present in the raio `a : b`, their average molar mass`=(axx16+bxx28)/(a+b)="20 g mol"^(-1)` (Given)
i.e., `16a+28b=20(a+b) or 4a+7b=5(a+b) or a=2b or (a)/(b)=(2)/(1)=2:1`
If the ratio is reversed, now the ratio `a:b=1:2`
`therefore" Average molar mass"=(1xx16+2xx28)/(1+2)=(16+56)/(3)=(72)/(3)="24 g mol"^(-1)`