Question

If a gaseous mixture is made by combining 3.02 g of Ar and 2.81 g of...

If a gaseous mixture is made by combining 3.02 g of Ar and 2.81 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?    (All three answers must be in units of atm)

PAr

PKr

PTotal

Homework Answers

Answer #1

As we know that the gases are ideal and we have the relation PV=nRT

P=total pressure of the system

V= volume of the system=2.5

n= total moles in the system

R= universal gas constant=0.0821

T= temperature of system=25+273=298 K

Atomic mass of Kr=83.8 g

Atomic mass of Ar=40 g

Moles of Kr=2.81/83.8=0.0335

Moles of Ar=3.02/40=0.0755

Hence total moles=0.0335+0.0755=0.109

Hence using the ideal gas equation

P=(0.109*0.0821*298)/2.5

P=1.067 atm

PAr= PTotal*Mole fraction of Ar=1.067*(0.0755/0.109)=0.739 atm

PKr= PTotal*Mole fraction of Kr=1.067*(0.0335/0.109)=0.328 atm

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