`1 L` of a gaseous mixture is effused in `5 min 11s`, while `1 L` of oxygen takes `10 min`. The gaseous mixture contains methane and hydrogen. Calculate
(`a`) The density of gaseous mixture.
(`b`) The percentage by volume of each gas in mixture.

Explanation

`{:(O_(2),"Mixture"),(d_(1)=16, d_(2)=?),(t_(1)=10xx60=600 s, t_(2)=5xx60+11=311 s):}`
Volume is same, so
`(t_(2))/(t_(1))=sqrt((d_(2))/(d_(1)))`
`(311)/(600)=sqrt((d_(2))/(16))`
`implies ((311)^(2))/((600)^(2))=(d_(2))/(16)`
`implies d_(2)=4.3`
`d_(H_(2))=1` and `d_(CH_(4))=8`. So
`d_(mix)=(d_(CH_(4))xxx+(100-x)d_(H_(2)))/(100)`
`4.3xx100=8x+(100-x)xx1`
`430=8x+100-x`
`330=7x`
`x=47.14%`
`%CH_(4)=47.14`, `%H_(2)=52.86`