Question

A mixture of three gases has a total pressure of 1,380 Torr at 298 K. The mixture is analyzed 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 torr (=356.04 mmHg)

Explanation:

Step 1: Data given

A mixture of three gases has a total pressure of 1,380 Torr at 298 K.

Moles of CO2 = 1.27 mol

Moles of CO = 3.04 mol

Moles of Ar = 1.50 Ar

Step 2: Calculate total number of moles

Total number of moles = Moles of CO2 + Moles of CO + Moles of Ar = 1.27 + 3.04 + 1.50 = 5.81 moles

Step 3: Calculate mol fraction Ar

Mol fraction Ar = 1.50/5.81 = 0.258

Step 4: Calculate partial pressure

1380 torr * 0.258 moles Ar = 356.04 torr = 356.04 mmHg

The partial pressure of Ar is 356.04 torr (=356.04 mmHg)