A mixture of three gases has a total pressure of 1,380 mm Hg 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

P (Ar) = 358.8 mmHg

Explanation:

Given dа ta:

Total pressure of gas = 1380 mmHg

Moles of CO₂ = 1.27 mol

Moles of CO = 3.04 mol

Moles of Ar = 1.50 mol

Partial pressure of Ar = ?

Solution:

Formula:

P (Ar) = [n₁/nt] Pt ... (1)

n₁ = moles of Ar

nt = total number of moles

Pt = total pressure

Now we will calculate the total number of moles.

nt = 1.27 mol + 3.04 mol + 1.50 mol

nt = 5.81 mol

Now we put the values in formula (1)

P (Ar) = [n₁/nt] Pt

P (Ar) = [1.50/5.81] 1380 mmHg

P (Ar) = 0.26 * 1380 mmHg

P (Ar) = 358.8 mmHg