A mixture of C H 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 C O 2 and H 2 O ( l ). C O 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 C H 4 in the original gas mixture?

Question

A mixture of C H 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 C O 2 and H 2 O ( l ). C O 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 C H 4 in the original gas mixture?, 19 21, 19 20, 17 18, 15 16

MARKS App · Accepted Answer

Let no. of moles of C H 4 present = n 1 mole Let no. of moles of C 2 H 2 present = n 2 mole CH 4 n 1 +2O 2 → CO 2 n 1 +2H 2 O C 2 H 2 n 2 + 5 2 O 2 → 2 CO 2 2n 2 +H 2 O T o t a l n o . o f m o l e s a t i n i t i a l = n 1 + n 2 T o t a l n o . o f m o l e s a t f i n a l = n 1 + 2 n 2 At constant V & T P 1 P 2 = N o . o f m o l e s a t i n i t i a l N o . o f m o l e s a t f i n a l 63 69 = n 1 + n 2 n 1 + 2 n 2 ⇒ 21 23 = n 1 + n 2 n 1 + 2 n 2 ⇒ n 2 n 1 = 2 19 ⇒ n 1 + n 2 n 1 = 21 19 ∴ n 1 n 1 + n 2 = 19 21 Alternative solutions CH 4 g + 2 O 2 g → CO 2 g + 2 H 2 O l 6 3 - x Torr 6 3 - x Torr C 2 H 2 g + 5 2 O 2 g → 2 CO 2 g + H 2 O l x Torr 2 x Torr ∑ P CO 2 = 6 3 - x + 2 x = 6 3 + x = 69 Torr ∴ x = 6 Torr X CH 4 = 6 3 - 6 6 3 = 5 7 6 3 = 1 9 2 1

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