Question

56g of nitrogen and 96g of oxygen are mixed isothermally and the mixture exerts a total pressure of 10 atm. The partial pressure of nitrogen

Answer 1

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,  

pA=xA×p

pB=xB×p

moles of nitrogen gas is nN2=56/28=2mol

moles of oxygen gas is nO2=96/32=3mol

Mole fraction of nitrogen is xN2=2/2+3=0.4

Mole fraction of oxygen will be xO2=1−0.4=0.6

Partial pressure of oxygen is pO2=0.6×10=6atm

Partial pressure of Nitrogen is pN2=0.4×10=4atm