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?

Question

Maximo Arroyo

Chemistry

29 October, 13:30

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?

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Jonathan Haney · 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