Solveeit Logo
Login

Question

Question: A mixture of CH<sub>4</sub> and C<sub>2</sub>H<sub>2</sub> occupied a certain volume at a total pres...

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 and H2O (l). 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?

A

1921\frac { 19 } { 21 }

B

1920\frac { 19 } { 20 }

C

1718\frac { 17 } { 18 }

D

1516\frac { 15 } { 16 }

Answer

1921\frac { 19 } { 21 }

Explanation

Solution

Let no. of moles of CH4 present = n1 mol. Let no. of moles of C2H2 present = n2 mol.

CH4 + 2O2 ¾® CO2 + 2H2O n1 n1 C2H2 + 52\frac { 5 } { 2 }O2 ¾® 2CO2 + H2O n2 2n2 Total no. of moles at initial = n1 + n2

Total no. of moles at final = n1 + 2n2

At constant V & T, n1+n2n1+2n2\frac { \mathrm { n } _ { 1 } + \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } + 2 \mathrm { n } _ { 2 } }

Ž 219\frac { 2 } { 19 } Ž n1+n2n1\frac { \mathrm { n } _ { 1 } + \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } }= 2119\frac { 21 } { 19 }

\ n1n1+n2\frac { \mathrm { n } _ { 1 } } { \mathrm { n } _ { 1 } + \mathrm { n } _ { 2 } } = 1921\frac { 19 } { 21 }