A container holds 22.4 litre of a gas at 1 atmospheric pressure and at 0°C. The gas consists of amixture of argon, oxygen and sulphur dioxide in which :(a) Partial pressure of SO2 = (Partial pressure 02) + (Partial pressure of Ar).(b) Partial pressure of O2 = 2 x partial pressure of Ar.Calculate the density of the gas mixture under these conditions.
Answers
Answer:
2.157 g/mol.
Explanation:
From the given question we have that the PSO2=PO2+PAr = 3PAr.
We also know that the partial pressure is given as
PSO2+PO2+PAr=1
And, 3PAr+2PAr+PAr =1
So, on equating boty of the equations we get that PAr =1/6, PO2=1/3 and PSO2=1/2.
From the formulae PV=nRT, the number od moles will be 0.9994.
Since, PAr will be the mole fraction(Ar)*Total pressure.
Since, the mole fraction of Ar =1/6
So, moles of Ar by the total moles will be equal to 1/6 .
Ar = 0.9994/6 = 0.16 moles.
O2 =0.9994/6 = 0.33 moles.
SO2 = 0.9994/6 = 0.499.
Since, the formulae of the moles will be given mass/molecular weight
We will get that the mass of Ar = 6.4, mass of O2 =10.56 and mass of SO2= 31.36.
Therefore, the total mass will be =48.32g.
Hence, the dennsity will be = Mass/volume which will be 48.32/22.4 which on solving we will get 2.157 g/mol.
Answer:
2.157
Explanation: