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Dalton's Law of Partial Pressures

A gaseous mix...

Question

A gaseous mixture containing $C O$ , $C H_{4}$ and $N_{2}$ gas has total volume of 40 $m L$ . This mixture is exploded with excess of oxygen, on cooling 30 $m L$ of contraction is observed and when this mixture is treated with aqueous $K O H$ a further contraction of 30 $m L$ is observed. Find the composition of the mixture in $m L$ .

C O = 20, C H_{4} = 10, N_{2} = 10

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C O = 15, C H_{4} = 15, N_{2} = 10

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C O = 10, C H_{4} = 15, N_{2} = 15

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C O = 20, C H_{4} = 15, N_{2} = 5

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Solution

The correct option is B $C O = 15, C H_{4} = 15, N_{2} = 10$
Let volume of $C O, C H_{4}, N_{2}$ be x, y, z $m L$ respectively.
On explosion with excess of $O_{2}$ the following reaction takes place:
$C O x m L (g) + \frac{1}{2} O_{2} (g) \to C O_{2} x m L (g)$
$C H_{4} y m L (g) + 2 O_{2} (g) \to C O_{2} y m L (g) + 2 H_{2} O (g) 2 y m L$
$N_{2}$ remains unreacted.
On cooling $H_{2} O$ gas liquifies hence, volume reduction of 30 $m L$ is observed.
$∴$ 2y = 30 $⟹$ y = 15
Volume of $C H_{4}$ = 15 $m L$
But volume of $C O_{2}$ = (x + y)
As we know $C O_{2}$ is absorbed on $K O H$ and volume reduction of 30 $m L$ is observed.
(x + y) = 30 $⟹$ x = 15 $m L$
Volume of $C O$ = 15 $m L$
$∵$ x + y+ z = 40 $m L$ $⟹$ z = 10 $m L$
Volume of $N_{2}$ = 10 $m L$

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Dalton's Law of Partial Pressures

Standard XII Chemistry

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