Question

In 2.00 min, 29.7 mL of He effuse through a small hole. Under the same conditions of pressure and temperature, 9.28 mL of a mixture of CO and CO2 effuse through the hole in the same amount of time. Calculate the percent composition by volume of the mixture. The effusion rate of a gas is proportional to its root-mean-square speed, which is related to its molar mass.

Answer 1

Answer : The percent composition by volume of mixture of
`CO` and
`CO_2` 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.

`R\propto \sqrt{(1)/(M)}`

And the relation between the rate of effusion and volume is :

`R=(V)/(t)`

or, from the above we conclude that,

`((V_1)/(V_2))^2=(M_2)/(M_1)` ..........(1)

where,

`V_1` = volume of helium gas = 29.7 ml

`V_2` = volume of mixture = 9.28 ml

`M_1` = molar mass of helium gas = 4 g/mole

`M_2` = molar mass of mixture = ?

Now put all the given values in the above formula 1, we get the molar mass of mixture.

`((29.8ml)/(9.28ml))^2=(M_2)/(4g/mole)`

`M_2=40.97g/mole`

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
`CO` be, 'x' and the mole fraction of
`CO_2` will be, (1 - x).

As we know that,

`\text{Average molar mass of mixture}=\text{Mole fraction of }CO`

`\text{Average molar mass of mixture}=(\text{Mole fraction of }CO* \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2* \text{Molar mass of } CO_2)`

Now put all the given values in this expression, we get:

`40.94g/mole=((x)* 28g/mole)+((1-x)* 44g/mole)`

`x=0.1894`

The mole fraction of
`CO` = x = 0.1894

The mole fraction of
`CO_2` = 1 - x = 1 - 0.1894 = 0.8106

The percent composition by volume of mixture of
`CO` =
`0.1894* 100=18.94\%`

The percent composition by volume of mixture of
`CO_2` =
`0.8106* 100=81.06\%`

Therefore, the percent composition by volume of mixture of
`CO` and
`CO_2` are, 18.94 % and 81.06 % respectively.