A mixture of `CH_(4)` and `C_(2)H_(2)` occupies a certain volume at a total pressure of 70.5 mm Hg. The sample is burnt, formed `CO_(2)` and `H_(2)O`. The `H_(2)O` is removed and the remaining `CO_(2)` is found to have pressure of 96.4 mmHg at the same volume and temperature as the original mixture. What mole fraction of the gas was `C_(2)H_(2)` ?

Explanation

Let the number of moles of `CH_(4)` and `C_(2)H_(2)` be x and y respectively.
`{:(CH_(4),+,C_(2)H_(2),+,O_(2),rarr,CO_(2) ,+,H_(2)O),("x moles",,"y moles",,,,,,):}`
Applying POAC for C atoms,
`1 xx` moles of `CH_(2) + 2 xx` moles of `C_(2)H_(2) = 1 xx` moles of `CO_(2)`
`x + 2y = "moles of " CO_(2)`
As no. of moles `alpha` pressure at const. temperature and volume.
`("no. of moles of " CH_(4) " and "C_(2)H_(2))/("no. of moles of " CO_(2)) = (70.5)/(96.4)`
or `(x+y)/(x+2y) = (70.5)/(96.4)`
`therefore (y)/(x+y)` = mole fraction of `C_(2)H_(2) =0.368`.