A mixture of carbon dioxide and argon gases at a total press

A mixture of carbon dioxide and argon gases, at a total pressure of 727 mm Hg, contains 8.06 grams of carbon dioxide and 4.46 grams of argon. What is the partial pressure of each gas in the mixture? mm Hg mm Hg 8 % - PAr

Solution

First we calculate the no. of moles of argon and carbon dioxide present in the mixture,

Molar mass of Argon = 40 g / mol

No. of moles of Argon = Mass of Argon given / Molar mass

No. of moles of Argon = ( 4.46 g ) / ( 40 g/mol ) = 0.112 mol

Molar mass of CO₂ = 44 g / mol

No. of moles of CO₂ = Mass of CO₂ given / Molar mass

No. of moles of CO₂ = ( 8.06 g ) / ( 44 g/mol ) = 0.183 mol

Now,

Mole fraction of Argon in mixture = Moles of Argon Present / Total no. of moles in mixture

Mole fraction of Argon in mixture = 0.112 / (0.112 + 0.183) = 0.380

Mole fraction of CO₂ in mixture = Moles of CO₂ Present / Total no. of moles in mixture

Mole fraction of CO₂ in mixture = 0.183 / (0.112 + 0.183) = 0.620

Partial Pressure of Argon = Mole fraction of Argon * Total Pressure

Partial Pressure of Argon = (0.380) * (727 mmHg) = 276.26 mmHg

Partial Pressure of CO₂ = Mole fraction of CO₂ * Total Pressure

Partial Pressure of CO₂ = (0.620) * (727 mmHg) = 450.74 mmHg