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Question

# 200 ml of gaseous mixture containing CO, CO2 and N2 on complete combustion in just sufficient amount of O2 showed a contraction of 40 ml . When the resulting gases were passed through KOH solution it reduces by 50 ml, then calculate the volume of VCO:VCO2:VN2 in the original mixture.

A
4 : 1 : 5
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B
2 : 3 : 5
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C
1 : 4 : 5
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D
1 : 3 : 5
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Solution

## The correct option is A 4 : 1 : 5Given, Volume of CO, CO2,N2 =200 mlOn combustion, CO2 remains intact and nitrogen does not react at low temperatures.2CO +O2→2CO2x x/2 xTotal Volume = 200 + x2Resulting mixture = N2 = y ml CO2 =200−y mlContradiction = Final - Initial = 200 +x2−200=x2 40=x2 Hence x=80 mlWhen mixture passes through KOH, only CO2 reacts.As volume reduces by 50% , 200−y=100 ml y=100 mlHence CO=80 ml, y=100 ml and CO2 = 20 mlVCO:VCO2:VN2 =4:1:5Therefore, option A is correct.

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