Graham’s Law of Diffusion calculation

 Graham’s law states that any gas's diffusion rate at constant temperature and pressure is directly proportional to the square root of its relative molecular mass or vapour density.

R₁/R₂ = √m₂/√m₁                    or t₂/t₁ = √m₂/√m₁

Example Questions

1.       If 200cm³ of hydrogen diffused through a porous pot in 30 seconds. How long will it take 400cm³ of oxygen to diffuse through the same pot? (H = 1, O = 16)

Solution

Before solving the problem, the first thing is to find how long it will take an equal volume (400cm³) of hydrogen to diffuse through the pot. This is because we are comparing oxygen and hydrogen.

Since 200cm³ of hydrogen = 30s

400cm³ will = 30/200 X 400 = 60s

        t₂/t₁ = √m₂/√m₁

        .t₁ = 60s

         .t₂ = ??

         M₁ = 2

          M₂ = 32

               60/.t₂ = √2/√32   

            .t² ₂  = 60x60 x 16

                   = 57600

            .t₂ = √57600 = 240sec

2.       Under the same conditions of temperature and pressure hydrogen diffuses 2 times as fast as a gas, X. Calculate the relative molecular mass of X (relative molecular = 2)

R of hydrogen = 2

R  of X  = 1

M of H =  2

1/2 = √2/√m

Square both sides to remove the square root

1/4 = 2/m_x                       M_x = 4x2 = 8

Relative molecular mass of X = 8

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