Question

A tank, containing O2 and H2, has a volume of 4.00 L and at total pressure is 24.44 atm. Calculate the mass of oxygen of which mole fraction is 0.4 in the tank at 25.0 °C .

Answer 1

SOLUTION.

Given

The mixure of oxygen and H2 is contains inside the tank with mole fraction of O2 is 0.4 and H2 is 0.6.

V	4 lit
T	25 °C
P	24.44 atm

Assuming gas mixure as ideal gas mixure .

For an ideal gas we have a relationship .

N*R*T = P*V  

R is ideal gas constant .

N is moles of mixure in the thanks

N = PV/RT = 24.44 atm *4 lit /{(0.08206 lit atm /mol K) *(273+25)K }

N = 4 moles

In tank there are total 4 moles .

We know that sum of all mole fraction of system componant is equal to Unity.

We have two componants system with O2 =0.4 and H2 =(1-0.4)= 0.6  

Mole fraction of O2 = mole of O2/total moles

0.4= moles of O2/(4)

Thus moles of O2 =4*0.4 moles =1.6 moles

Oxygen mol wt =32 gm/mol

Mass of O2= 32 gm/mol *(1.6 moles)

Mass of O2 = 32*1.6 =51.2 gms.

ANSWER ,

Mass of oxygen inside the tank is 51.2 gram.

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