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) ?

A mixture of CH_(4) and C_(2)H_(2) occupies a certain volume at a tota

1.	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) ?
Answer» Solution :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`.