Answer
Partial pressure of Ar = 0.372 atm
Explanation
Given:
Total pressure = 1345 mmHg = 1.7697 atm
amounts:
O2 = 48% = 0.48 g
Ar = 25% = 0.25 g
N2 = 15% = 0.15 g
F2 = 12% = 0.12 g
Required: To find the partial pressure of Ar in the mixture
Solution
Ptot = P1 + P2 + P4 + ...
Ptot = ntotRT/V
Step 1: Find the moles of Ar and all other gasses
Ar:
n = m/M
n = 0.25g/39.95g/mol
n = 6.258x10^-3 mol
O2:
n = m/M
n = 0.48g/31.998g/mol
n = 0.015 mol
N2:
n = 0.15g/28.01g/mol
n = 5.355x10^-3 mol
F2:
n = 0.12g/37.997g/mol
n = 3.158x10^-3 mol
Step 2: Find the partial pressure of Argon (Ar)
Total number of moles = 6.258x10^-3 mol + 0.015 mol + 5.355x10^-3 mol + 3.158x10^-3 mol
= 0.0298 mol
P of Ar = (n of Ar/ n total) x P total
Pressure of Ar = (6.258x10^-3 mol/0.0298 mol) x 1.7697 atm
Pressure of Ar = 0.372 atm
