Question

i have a mixture of gases with a total pressure of 1345 mm Hg. i know that mixture is 48 percent O2, 25 Percent Ar, 15 percent N2 and 12 percent F2. Find the partial pressure of Ar in this mixture (in atm)

Answer 1

Partial pressure of Ar = 0.372 atm

Step-by-step 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