Amixture 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?

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)

it is b that is the answer.