Question

A gaseous mixture of O2 and N2 contains 36.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 585 Torr ?

Answer 1

let total mass be 100 g

mass of N2 = 36.8 g

mass of O2 = 100 - 36.8 = 63.2 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

= 36.8/28.02

= 1.3133

n(O2) = mass of O2/molar mass of O2

= 63.2/32.0

= 1.975

n(N2),n1 = 1.3133 mol

n(O2),n2 = 1.975 mol

Total number of mol = n1+n2

= 1.3133 + 1.975

= 3.2883 mol

Partial pressure of each components are

p(N2),p1 = (n1*Ptotal)/total mol

= (1.3133 * 585)/3.2883

= 234 torr

p(O2),p2 = (n2*Ptotal)/total mol

= (1.975 * 585)/3.2883

= 351 torr

Answer:

partial pressure of N2 = 234 torr

partial pressure of O2 = 351 torr