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12 mol gaseous mixture of an alkane and ...

12 mol gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing 20% v/v `O_(2)` and rest `N_(2)` ) for complete combustion at 200K . After combustion when gaseous mixture is passes through KOH solution it shows volume contraction of 36 ml.
Calculate mol % of oxygen which is converted into `H_(2)O`

A

0.3684

B

0.7368

C

0.2061

D

0.2563

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The correct Answer is:
A
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Knowledge Check

  • 12 mol gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing 20% v/v O_(2) and rest N_(2) ) for complete combustion at 200K . After combustion when gaseous mixture is passes through KOH solution it shows volume contraction of 36 ml. Formula of alkane is:

    A
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    B
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    C
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    D
    `C_(4)H_(10)`

  • 12 mol gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing 20% v/v O_(2) and rest N_(2) ) for complete combustion at 200K . After combustion when gaseous mixture is passes through KOH solution it shows volume contraction of 36 ml. Mole fraction of CO_(2) in final gaseous sample:

    A
    `(6)/(51)`
    B
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    C
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  • 12 ml of a mixture of alkane and alkene having same number of carbon atoms requires exactly 57 ml of oxygen for complete combustion. The name of hydrocarbons if CO_2 formed is 36 ml are

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    B
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    C
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