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Question

1 litre of a mixture of CO and CO2 is taken. This mixture is passed through a tube containing red-hot charcoal. The volume now becomes 1.6 liters. The volumes are measured under the same conditions. Find the composition of the mixture by volume.

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Solution

PV=nRTVα1mHere,V=volumem=molecularweightVCO+VCO2=1.6L(i)VCOVCO2=mCO2mCO=4428VCO=117VCO2Onputtingthevalueof(1)weget117VCO2+VCO2=1.6LVCO2=1.6×718=11.218=0.62LandVCO=117VCO2=0.98.Hence,VCO2=0.62LandVCO=0.98L

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