Question

A mixture of $CH _{4}$ and $C _{2} H _{2}$ occupied a certain volume at a total pressure equal to 63 torr. The same gas mixture was burnt to $CO _{2}$ and $H _{2} O (1)$. $CO _{2}( g )$ 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 $CH _{4}$ in the original gas mixture?

• A. $\frac{19}{21}$
• B. $\frac{19}{20}$
• C. $\frac{17}{18}$
• D. $\frac{15}{16}$

Answer 1

Answer: (A)

Let no. of moles of $CH _{4}$ present $= n _{1}$ mole

Let no. of moles of $C _{2} H _{2}$ present $= n _{2}$ mole

$\underset{n_{1}}{ CH _{4}}+2 O _{2} \rightarrow \underset{ n _{1}}{ CO _{2}}+2 H _{2} O$

$C _{2} H _{2}+\frac{5}{2} O _{2} \rightarrow 2 CO _{2}+ H _{2} O$

Totalno. of molesatinitial $= n _{1}+ n _{2}$

Totalno. of molesatfinal $= n _{1}+2 n _{2}$

At constant $V \& T$

$\frac{P_{1}}{P_{2}}=\frac{\text { No.ofmolesatinitial }}{\text { No. ofmolesatfinal }}$

$\frac{63}{69}=\frac{ n _{1}+ n _{2}}{ n _{1}+2 n _{2}} \Rightarrow \frac{21}{23}=\frac{ n _{1}+ n _{2}}{ n _{1}+2 n }$

$\Rightarrow \frac{ n _{2}}{ n _{ l }}=\frac{2}{19} \Rightarrow \frac{ n _{1}+ n _{2}}{ n _{1}}=\frac{21}{19}$

$\therefore \frac{ n _{1}}{ n _{1}+ n _{2}}=\frac{19}{21}$