`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

Step I. Calculation of density of gaseous mixutre.
According t available data :
Time taken for effusion by `O_(2) (t_(1)) = 10 xx 60 = 600 s`
Time taken for effusion by gaseous mixture `(t_(2)) = 5 xx 60 + 11 = 311s `
Density of oxygen `d_((O_(2))) = 16`
The density of gaseous mixture `d_(("mix"))` can be calculated by Graham's law of effusion.
`t_(2)/(t_(1)) = sqrt((d_("mix"))/(d_((O_(2)))))` or `(311)/(600) = sqrt((d_(("mix")))/(16))`
On squring both sides, `d_(("mix")) = ((311)^(2) xx 16)/((600)^(2)) = 4.3`
Step II. Calculation of percentage by volume of each gas in mixture.
Let the percentage of methane `(CH_(4)) = x`
`:.` the percentage of hydrogen `(H_(2)) = 100 - x`
According to the law of mixtures,
`d_(("mix")) = (x xx d_((CH_(4))) + (100-x) xx d_(H_(2)))/(100)`
`4.3 = (x xx 8 + (100 - x) xx 1)/(100) = (8x + 100 -x )/(100)`
`7 x = 430 - 100 = 330` or `x = (330)/(7) = 47.14`
Percentage of `CH_(4) = 47.14`
`:.` Percentage of `H_(2) = 100 - 47.14 = 52.86`