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.

$R\propto \sqrt{\frac{1}{M}}$

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

$R=\frac{V}{t}$

or, from the above we conclude that,

$(\frac{V_1}{V_2})^2=\frac{M_2}{M_1}$            ..........(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.

$(\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}$

$M_2=40.97g/mole$

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,

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

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

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

$40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)$

$x=0.1894$

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 $CO$ = $0.1894\times 100=18.94\%$

The percent composition by volume of mixture of $CO_2$ = $0.8106\times 100=81.06\%$

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