Calculate the mole fraction of N2 and O2 in the mixture where 56 gm N2 mixed with 64 gm O2.
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Raoult's Law states that the partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.
In a mixture of liquids of A and B,
p
A
=x
A
×p
p
B
=x
B
×p
moles of nitrogen gas is n
N
2
=56/28=2 mol
moles of oxygen gas is n
O
2
=96/32=3 mol
Mole fraction of nitrogen is x
N
2
=2/(2+3)=0.4
Mole fraction of oxygen will be x
O
2
=1−0.4=0.6
Partial pressure of oxygen is p
O
2
=0.6×10=6 atm
Partial pressure of Nitrogen is p
N
2
=0.4×10=4 atm
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