Question

A mixture of gas contains 0.215g Ar and 1.95g Xe and has a total pressure of 2.33atm. What are the partial pressures of each gas?

Answer 1

Molar mass of Ar = 39.95 g/mol

Molar mass of Xe = 131.3 g/mol

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

= 0.215/39.95

= 0.0054

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

= 1.95/131.3

= 0.0149

n(Ar),n1 = 0.0054 mol

n(Xe),n2 = 0.0149 mol

Total number of mol = n1+n2

= 0.0054 + 0.0149

= 0.0202 mol

Partial pressure of each components are

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

= (0.0054 * 2.33)/0.0202

= 0.620 atm

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

= (0.0149 * 2.33)/0.0202

= 1.71 atm

partial pressure of Ar = 0.620 atm

partial pressure of Xe = 1.71 atm