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Question

10 ml mixture of CO,CH4 and N2, exploded with an excess of oxygen, gave a contraction of 6.5 ml. There was a further contraction of 7ml when the residual gas was treated with KOH. Volume of CO,CH4 and N2 respectively, is :

A
5ml, 2ml, 3ml
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B
2ml, 6ml, 4ml
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C
1ml, 8ml, 9ml
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D
4ml, 3ml, 5ml
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Solution

The correct option is A 5ml, 2ml, 3ml
Solution:
Let the volumes of CO,CH4, and N2 in the mixture be x ml, y ml, and (10-x-y) ml respectively. We ignore the volume of water formed in the combustion as it is a liquid and its volume is small. Nitrogen combusts only at very very high temperatures. So it remains unreactive.

2CO+O22CO2 CH4+2O2CO2+2H2O
x ml x/2 ml x ml y ml 2 y ml y ml (liquid)

The volume of oxygen used: 2y+x2ml
Total volume of all gases before combustion =10+2y+x2ml
After combustion the total volume is: 10 ml
as CO2:x+y N2:10xy

Reduction in volume: 10+2y+x210=2y+x2=6.5ml
4y+x=13ml --- (1)

When the mixture of CO2+N2 passes over KOH, all of CO2 is absorbed. Only N2 remains. Then the reduction of 7 ml corresponds to that of CO2.
x+y=7ml --- (2)

Solving the two equations, x = 5 ml and y = 2 ml

Volumes of CO=5ml, CH4=2ml , N2=3ml.


Hence A is the correct option

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