Question

100 mL of hydrogen was confined in a diffusion tube and exposed to air and at equilibrium, a volume of 26.1 mL of air was measured in the tube. Again, when 100 mL of CO2 was placed in the same tube and exposed to air, 123 mL of air was measured in the tube at the equilibrium. Find the molecular weight of CO2.

Answer 1

Answer : The molar mass of a $CO_2$ is, 44.4 g/mole

Solution :

According to the Graham's law, the rate of diffusion 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 diffusion 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)

First we have to calculate the molecular weight of air.

$V_1$ = volume of hydrogen gas = 100 ml

$V_2$ = volume of air = 26.1 ml

$M_1$ = molar mass of hydrogen gas  = 2 g/mole

$M_2$ = molar mass of air = ?

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

$(\frac{100ml}{26.1ml})^2=\frac{M_2}{2g/mole}$

$M_2=29.36g/mole$

Now we have to calculate the molecular weight of $CO_2$.

$V_1$ = volume of air = 123 ml

$V_2$ = volume of $CO_2$ gas = 100 ml

$M_1$ = molar mass of air   = 29.36 g/mole

$M_2$ = molar mass of $CO_2$ gas = ?

Now put all the given values in the above formula 1, we get the molar mass of $CO_2$ gas.

$(\frac{123ml}{100ml})^2=\frac{M_2}{29.36g/mole}$

$M_2=44.4g/mole$

Therefore, the molar mass of a $CO_2$ is, 44.4 g/mole