Answer :
Answer : The percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
[tex]R\propto \sqrt{\frac{1}{M}}[/tex]
And the relation between the rate of effusion and volume is :
[tex]R=\frac{V}{t}[/tex]
or, from the above we conclude that,
[tex](\frac{V_1}{V_2})^2=\frac{M_2}{M_1}[/tex] ..........(1)
where,
[tex]V_1[/tex] = volume of helium gas = 29.7 ml
[tex]V_2[/tex] = volume of mixture = 9.28 ml
[tex]M_1[/tex] = molar mass of helium gas = 4 g/mole
[tex]M_2[/tex] = molar mass of mixture = ?
Now put all the given values in the above formula 1, we get the molar mass of mixture.
[tex](\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}[/tex]
[tex]M_2=40.97g/mole[/tex]
The average molar mass of mixture = 40.97 g/mole
Now we have to calculate the percent composition by volume of the mixture.
Let the mole fraction of [tex]CO[/tex] be, 'x' and the mole fraction of [tex]CO_2[/tex] will be, (1 - x).
As we know that,
[tex]\text{Average molar mass of mixture}=\text{Mole fraction of }CO[/tex]
[tex]\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)[/tex]
Now put all the given values in this expression, we get:
[tex]40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)[/tex]
[tex]x=0.1894[/tex]
The mole fraction of [tex]CO[/tex] = x = 0.1894
The mole fraction of [tex]CO_2[/tex] = 1 - x = 1 - 0.1894 = 0.8106
The percent composition by volume of mixture of [tex]CO[/tex] = [tex]0.1894\times 100=18.94\%[/tex]
The percent composition by volume of mixture of [tex]CO_2[/tex] = [tex]0.8106\times 100=81.06\%[/tex]
Therefore, the percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.