Question

If a gaseous mixture is made by combining 3.99 g of Ar and 4.24 g of Kr in an evacuated 2.50 L container at 25.0 °C, what are the partial pressures of each gas and what is the total pressure exerted by the gaseous mixture?

Answer 1

Given

Mass of Ar = 3.99 g

Mass of Kr= 4.24 g

Volume = 2.50 L

T = 25.0 deg C = 25.0+273.15 = 298.15 K

Calculation of moles of each

Moles of Ar = 3.99 g / molar mass of Ar , 39.948 g per mol

= 0.0999 mol Ar

Moles of Kr = 4.24 g / 83.8 g per mol

= 0.0506 mol

Total moles = moles of Ar + moles of Kr

= 0.0999 mol + 0.0506 mol = 0.1505 mol

Calculation of total pressure

pV = nRT

p = nRT / V

here , n = moles , R is gas constant = 0.08206 L atm (Kmol)^-1

Lets plug the values to get total pressure

p = 0.1505 mol x 0.08206 L atm (Kmol) x 298.15 K/ 2.50 L

= 1.47 atm

Partial pressure of Ar = mol fraction x total pressure

= (mol Ar/ Total moles ) x 1.47 atm

= (0.0999 mol / 0.1505 mol ) x 1.47 atm

= 0.976 atm

Parial pressure of Kr

= (0.0506 / 0.1505 ) x 1.47 atm

= 0.494 atm