A 1:1 mixture (by weight) of hydrogen and helium is enclosed in a one-litre flask at 0∘C. Assuming ideal behaviour, the partial pressure of helium is found to be 0.42atm, then the concentration of hydrogen would be:
A
0.0375 M
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B
0.028 M
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C
0.0562 M
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D
0.0187 M
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Solution
The correct option is A0.0375 M Let the weights of H2 and He in the mixture be x each.
Then the number of moles of H2=x2
and the number of moles of He=x4
Pressure in the flask =P=nRTV P=(x2+x4)×0.0821×2731 ⇒P=16.81x
Partial pressure of He=0.42atm ⇒x/4(x2+x4)×P=0.42 ⇒x/43x/4×16.81x=0.42 ⇒x=0.075 ∴ concentration of H2=x/2V=0.075/21=0.0375M