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  • Solve it, A 40 ml of a mixture of H2 and O2 at 18 ºC and 1 atm pressure was sparked so that the formation of water was complete. The remaining pure gas had a volume of 10 ml at 18ºC and 1 atm pressure. If the remaining gas was H2, the mole

A 40 ml of a mixture of H2 and O2 at 18 ºC and 1 atm pressure was sparked so that the formation of water was complete. The remaining pure gas had a volume of 10 ml at 18ºC and 1 atm pressure. If the remaining gas was H2, the mole fraction of H2 in the 40 ml mixture is:

  • Option 1)

    0.75

  • Option 2)

    0.5

  • Option 3)

    0.65

  • Option 4)

    0.85

 

Answers (1)

best_answer

As we learnt in 

Partial Pressure of a gas in gaseous mixture -

Partial Pressure = (Total Pressure of Mixture) \times (mole fraction of gas)

 

-

\\H_{2}+1/2O_{2}\rightarrow H_{2}O{(l)}\\ a\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: b\ \: \: \: \: \: \: \: \: \: \: \: \: 0\\ a-2b\ \: \: \: \: 0\ \: \: \: \: \: \: \: \: \: \: \: \: b

Reaction is studied at constant P &T.

a+b = 40 \: \: \: \: a-2b = 10

a = 30 ml \: \: \: \: \: b = 10ml

mole fraction of H2 =  volume fraction of H2 =30/40 = 0.75.

 


Option 1)

0.75

Correct

Option 2)

0.5

Incorrect

Option 3)

0.65

Incorrect

Option 4)

0.85

Incorrect

Posted by

Plabita

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