Question

Calculate the mole fraction of N2 and O2 in the mixture where 56 gm N2 mixed with 64 gm O2.​

Answer 1

Answer:

Explanation:

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