A gas mixture contains `50%` helium and `50%` methane by volume. What is the percent by weight of methane in the mixture.

Options

A. 0.1997
B. 0.205
C. 0.5
D. 0.8003

Answer

D

Explanation

`PV = nRT`
`PV = (w)/(M) RT`
`V = (w)/(M) (RT)/(P)`
Here R, T and P are constants
`:. V_(He) prop (w(He))/(M(He))`
and `V_(CH_(4)) prop (w(CH_(4)))/(M(CH_(4)))`
But `V_(He) = V_(CH_(4))` (Given)
`(w(He))/(M(He)) = (w(CH_(4)))/(M(CH_(4)))`
`(w(He))/(4) = (w(CH_(4)))/(16)`
`(w(He))/(w(CH_(4)))= (1)/(4)`
`w(CH_(4)) = 4 w (He)`
Wt. % of `CH_(4) = (w(CH_(4)) xx 100)/(w(CH_(4)) + w(He))`
`= (4 xx 100)/(5) = 80%`