Question

A mixture of three gases has a pressure at 298 K of 1380 mm Hg. The mixture is analysed and is found to contain 1.27 mol CO2, 3.04 mol CO, and 1.50 mol Ar. What is the partial pressure of Ar?

Answer 1

Answer:

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

Explanation:

Step 1: Data given

A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K

Moles of CO2 = 1.27 moles

Moles of CO = 3.04 moles

Moles of Ar = 1.50 moles

Step 2: Calculate total number of moles

Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles

Step 3: Calculate mol fraction Ar

Mol fraction Ar = 1.50 mol/5.81 mol = 0.258

Step 4: Calculate partial pressure

1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)