A mixture of CH₄ and C₂H₂ occupied a certain volume at a total pressure equal to 63 torr. The same gas mixture was burnt to CO₂(g) and H₂O(1). The CO₂(g) alone was
collected in the same volume and at the same temperature, the pressure was found to be 69 torr.

What was the mole fraction of CH₄ in the original gas mixture?

Let no. of moles of CH₄ present: n, mol. 

Let no. of moles of ${\mathrm{C}}_2{\mathrm{H}}_2$ present = n₂ mol.

$\therefore \left(n_1+n_2\right)=63\text{ torr (1)}$

$\begin{array}{l}{\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ n_1\mathrm{mol}\to n_1\mathrm{mol}\\ {\mathrm{C}}_2{\mathrm{H}}_2+3{\mathrm{O}}_2\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ n_2\mathrm{mol}\to 2n_2\mathrm{mol}\end{array}$

.'. After combustion total no. of moles

$=\left(n_1+2n_2\right)=69\text{ torr (2)}$

$\therefore n_2=6\text{ torr and }{n}_1=57\text{ torr }$

$\therefore \text{ Mole fraction of }{\mathrm{CH}}_4\text{ in the original gas mix }=\frac{57\text{ torr }}{63\text{ torr }}=\frac{19}{21}$