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

This question is based on Graham's law of diffusion

according to him, rate of diffusion of gas is inversely proportional to square root of molar mass of that gas.

and we know, rate of diffusion = volume of diffused gas /time taken to diffuse

so, volume of diffused gas $\propto$ 1/√{molar mass of gas}

or, $\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}$

case 1 : volume of hydrogen after diffusion = 100mL

volume of air after diffusion = 26.1mL

let molar mass of air is M

molar mas of H2 = 2g/mol

so, 100mL/26.1mL = √{M/2}.....(1)

case 2 : volume of air after diffusion = 123mL

volume of CO2 after diffusion = 100mL

Let molar mass of CO2 is X

so, 123/100 = √{M/X}......(2)

from equations (1) and (2),

123/26.1 = √{X/2}

or, 4.71 = √{X/2}

or, X ≈ 44 g/mol

hence, molar mass of CO2 is 44g/mol.