Question

In a mixture of helium and chlorine, occupying a volume of 14.6 l at 871.7 mmhg and 28.6oc, it is found that the partial pressure of chlorine is 355 mmhg. what is the total mass of the sample?

Answer 1

First, let's find the total moles.

PV=nRT
(871.7 mmHg)(1 atm/760 mmHg)(14.6 L) = n(0.0821 L-atm/mol-K)(28.6 +273)
Solving for n,
n = 0.676 moles

Assuming ideal gas behavior, the pressure fraction is also equal to mole fraction.

Mole fraction of Chlorine = 355/871.7 = 0.407
Mole fraction of Helium = 1 - 0.407 = 0.593

Knowing Chlorine to be 35.45g/mol and Helium to be 4 g/mol.
Mass of Chlorine = (0.407)(0.676)(35.45 g/mol) = 9.75 g
Mass of Helium = (0.593)(0.676)(4 g/mol) = 1.6 g
Total Mass = 9.75+1.6 = 11.35 g