`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 `CO_(2)` 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 `CO_(2)`.

Explanation

In the first case, when `100 mL` of `H_(2)` is diffused, the volume of air diffused was `26.1 mL`. Thus
`(Rate of diffusion of H_(2))/(Rate of diffusion of air)=(100)/(26.1)` …(`i`)
`(Rate of diffusion of CO_(2))/(Rate of diffusion of air)=(100)/(123)` ...(`ii`)
From equation (`i`) and (`ii`), we get
`(r_(H_(2)))/(r_(CO_(2)))=(123)/(261)=sqrt((M_(CO_(2)))/(M_(H_(2))))`
`Molecular of weight of CO_(2)=(123xx123)/(26.1xx26.1)=2=44.42`