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Question

# 12 ml of a gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing 20% by volume of O2 and rest N2) for complete combustion at room temperature. After combustion when the gaseous mixture is passed through a KOH solution, it shows a volume contraction of 36 ml. Find the true statement(s) in the following options:

A
The alkane is propane
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B
The alkene is butene
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C
Mole fraction of CO2 in the final gaseous mixture is 322
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D
Liquid water formed is 42 ml
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Solution

## The correct option is C Mole fraction of CO2 in the final gaseous mixture is 322Let the alkane be CnH2n+2 and alkene be CnH2n and let x ml be the volume of the alkane and 12−x ml be the volume of alkene. CnH2n+2(g)+3n+12O2(g)→nCO2(g)+(n+1)H2O(l) CnH2n(g)+3n2O2(g)→nCO2(g)+nH2O(l) Volume of CO2 produced by combustion of alkane = nx ml Volume of CO2 produced by combustion of alkene = n(12−x) ml The contraction of gaseous mixture in KOH is due to CO2, i.e. vol of CO2 is 36 ml. So, nx+n(12−x)=36⇒n=3 a) The alkane is propane. b) The alkene is propene. According to the question, N2 in 285 ml air =285×0.80=228 ml c) So, mole fraction of CO2 in final gaseous mixture =3636+228=36264=322 d) The volume of oxygen is 285×20100=57 ml ∴x×3n+12+(12−x)×3n2=57 Putting, n=3 we get, x=6. So, 6 ml of propane and 6 ml of propene are present in the mixture. So moles of liquid water formed, 6 mL×(n+1+n) mol22.4 L=6×722.4 mmol Therefore mass of liquid water will be, 6×722.4 mmol×18 g/mol=33.75 mg So, volume of liquid water will be 0.03375 ml because the density of water at room temperature is 1 g/mL.

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