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 : The percent composition by volume of mixture of and 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.

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

or, from the above we conclude that,

           ..........(1)

where,

= volume of helium gas = 29.7 ml

= volume of mixture = 9.28 ml

= molar mass of helium gas  = 4 g/mole

= molar mass of mixture = ?

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

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

As we know that,

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

The mole fraction of = x = 0.1894

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

The percent composition by volume of mixture of =

The percent composition by volume of mixture of =

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

the answer is 0.8m.

by: nicholas