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?

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}

Tardigrade Admin · Accepted Answer

Let no. of moles of CH 4 present = n 1 mole Let no. of moles of C 2 H 2 present = n 2 mole undersetn1 CH 4+2 O 2 arrow underset n 1 CO 2+2 H 2 O C 2 H 2+(5/2) O 2 arrow 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 (P1/P2)=( text No.ofmolesatinitial / text No. ofmolesatfinal ) (63/69)=( n 1+ n 2/ n 1+2 n 2) ⇒ (21/23)=( n 1+ n 2/ n 1+2 n ) ⇒ ( n 2/ n l )=(2/19) ⇒ ( n 1+ n 2/ n 1)=(21/19) ∴ ( n 1/ n 1+ n 2)=(19/21)

1921

1920

1718

1516

$\frac{19}{21}$

$\frac{19}{20}$

$\frac{17}{18}$

$\frac{15}{16}$