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Question

Total pressure of a mixture of H2 and O2 is 1.0 bar. The mixture is allowed to react to form water, which is completely removed to leave only pure H2 at a pressure of 0.34 bar. Assuming ideal gas behaviour and that all pressure measurements were made under the same temperature and volume conditions, the composition of the original mixture is:

A
XH2XO20.500.50
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B
XH2XO20.220.78
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C
XH2XO20.780.22
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D
XH2XO20.350.65
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Solution

The correct option is C XH2XO20.780.22
Initially H2 and O2 both are present
So in PV=nRT,
nH2+nO2=n(total)=PVRT=1×VRT=VRT
After reaction is complete, only H2 (unreacted ) is left
Moles of unreacted H2=PVRT=0.34VRT
Thus, nH2+nO2+unreacted nH2=VRT
nH2+nO2=VRT0.34VRT
nH2+nO2=0.66VRT --- (i)
Also, 2H2(g)+O2(g)2H2O(l)
Moles of H2=2×moles of O2
so, nH2=2nO2
On putting this value in (i), we get:
3nO2=0.66VRT
nO2=0.66V3RT
XO2=nO2ntotal=0.66V/3RTV/RT=0.22
XH2=10.22=0.78

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