Question

If a gaseous mixture is made by combining 4.15 g Ar4.15 g Ar and 1.73 g Kr1.73 g Kr in an evacuated 2.50 L container at 25.0 ∘C,25.0 ∘C, what are the partial pressures of each gas, ?ArPAr and ?Kr,PKr, and what is the total pressure, ?total,Ptotal, exerted by the gaseous mixture?

?Ar=PAr=

atmatm

?Kr=PKr=

atmatm

?total=Ptotal=

atm

Answer 1

1)

Molar mass of Ar = 39.95 g/mol

mass(Ar)= 4.15 g

use:

number of mol of Ar,

n = mass of Ar/molar mass of Ar

=(4.15 g)/(39.95 g/mol)

= 0.1039 mol

Given:

V = 2.5 L

n = 0.1039 mol

T = 25.0 oC

= (25.0+273) K

= 298 K

use:

P * V = n*R*T

P * 2.5 L = 0.1039 mol* 0.08206 atm.L/mol.K * 298 K

P = 1.0163 atm

Answer: 1.02 atm

2)

Molar mass of Kr = 83.8 g/mol

mass(Kr)= 1.73 g

use:

number of mol of Kr,

n = mass of Kr/molar mass of Kr

=(1.73 g)/(83.8 g/mol)

= 2.064*10^-2 mol

Given:

V = 2.5 L

n = 0.0206 mol

T = 25.0 oC

= (25.0+273) K

= 298 K

use:

P * V = n*R*T

P * 2.5 L = 0.0206 mol* 0.08206 atm.L/mol.K * 298 K

P = 0.2015 atm

Answer: 0.202 atm

3)

Ptotal = p(Ar) + p(Kr)

= 1.02 atm + 0.202 atm

= 1.22 atm

Answer: 1.22 atm