Question

If a gaseous mixture is made by combining 2.88 g of Ar and 1.47 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

moles of Ar = mass of Ar / molar mass of Ar
= 2.88/39.948
= 0.0721 mol

moles of Kr = mass of Kr / molar mass of Kr
= 1.47/83.798
= 0.0175 mol

total mol = n1+n2
= 0.0721 + 0.0175 mol
=0.0896 mol

T = 25 oC = 298 K
usE:
P*V = n*R*T
P*2.50 = 0.0896*0.0821* 298
P = 0.877 mol

Answer: total pressure = 0.877 atm

Partial pressure of Ar = moles of Ar * total pressure / total moles
= 0.0721*0.877/0.0896
= 0.706 atm   

partial pressure of Kr = total pressure - Partial pressure of Ar
= 0.877 atm - 0.706 atm
= 0.171 atm