A mixture of CO and CO2 is found to have a density of 1.50g/litre at 30°C and 730 mm. What is the composition of the mixture?
Answers
Answer:
For a mixture of CO and CO2, d = 1.50 g/litre
P = 730 / 760 atm, T = 303K
PV = w / m RT; PV w / Vm RT
730 / 760 = 1.5 / m × 0.0821 × 303 = ∴ 38.85
i.e. molecular weight of mixture of CO and CO2 = 38.85
Let % of mole of CO be a in mixture then
Average molecular weight = a × 28 + (100 – a) 44 / 100
38.85 = 28a + 4400 – 44a / 100
a = 32.19
Mole % of CO = 32.19
Mole % of CO2 = 67.81
Explanation:
We want to determine the composition of a mixture of CO and CO2 which has a density of 1.5 g/litre at a temperature of 30o C and a pressure of 730 torr.
The ideal gas equation is PV=nRT, where P,V,n,R and T are the pressure, volume, number of moles, the gas constant and the absolute temperature respectively.
The density of a gas is, ρ=massvolume.
The mass of the gas is nM, where M is the mass of one mole of gas, which is equal in magnitude to the molecular weight of the gas.
⇒ ρ=nMV=MPRT.
For CO,M=28.011 and for CO2,M=44.010.
It is given that P=730 torr and T=30o C =303 K.
The gas constant, R=62.364 L torr/mol K.
⇒ ρCO=28.011×73062.364×303=1.082 g/litre.
⇒ ρCO2=44.010×73062.364×303=1.700 g/litre.
Let the fraction of CO be f⇒ the fraction of CO2 is 1−f.
According to Dalton’s law, in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
⇒1.082f+1.700(1−f)=1.5⇒0.618f=0.2.
⇒f=0.20.618=0.324⇒1−f=0.676.
⇒ The mixture contains 32.4%CO and 67.6%CO2, or, approximately one-third CO and two-third CO2.