Question

A gaseous mixture of O2 and N2 contains 40.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 805 mmHg ?

Answer 1

Let total mass of mixture be 100 g
Then mass of N2 = 40.8 g
Mass of O2 = 100 - mass of N2
= 100 - 40.8
= 59.2 g

Molar mass of N2 = 28.02 g/mol


Molar mass of O2 = 32 g/mol

n(N2) = mass of N2/molar mass of N2
= 40.8/28.02
= 1.4561

n(O2) = mass of O2/molar mass of O2
= 59.2/32
= 1.85

n(N2),n1 = 1.4561 mol
n(O2),n2 = 1.85 mol

Total number of mol = n1+n2
= 1.4561 + 1.85
= 3.3061 mol

Partial pressure of each components are
p(N2),p1 = (n1*Ptotal)/total mol
= (1.4561 * 805)/3.3061
= 355 mmHg

p(O2),p2 = (n2*Ptotal)/total mol
= (1.85 * 805)/3.3061
= 450 mmHg

Answer: 450 mmHg