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Graham’s Law of Diffusion

Graham’s law of diffusion states that the rate of diffusion of a gas at a given temperature is inversely proportional to the square root of its molecular mass, m.

I.e. R α 1/√m

Also, vapour density (ρ) = molar mass (g)/molar volume (v)

Molar mass = vapor density (ρ) x molar volume (v)

Therefore, R α 1/√ρv

For two gases with molecular masses m₁ and m₂ ; and rates of diffusion R₁ and R₂:

R₁/R₂ = √m₂/√m₁ – (1)

R₁/R₂ = √ρ₂v₂ /√ρ₁v₁ – (2)

Also, rate of diffusion = volume/time I.e. the greater the rate, the shorter the time.

Comparing the rates of diffusion of equal volumes of two gases in times t₁ and t₂, we have R₁ = V/t₁ and R₂ = V/t₂

Note:

- The lighter or less dense a gas is, the greater is its rate of diffusion.

- If the gases are at the same temperature and pressure, they must have the same molar volumes. Therefore,

R₁/R₂ = √ρ₂ /√ρ₁ – (3)

R₁/R₂ = t₂/t₁ – (4)

Therefore, R₁/R₂ = t₂/t₁ = √m₂ /√m₁ = (5)

By the above last equation, equation (5), you could determine the value of any of the parameters by establishing an equation with those whose values are given.

Note:

- The vapour density (v.d.) of a gas is equal to half its relative molecular mass(r.m.m.).

I.e. r.m.m. = 2 x v. d.

30 cm³ of a gas, the empirical formula of which was CH₃, diffuses through a porous partition in 45.2 s. 30 cm³ of hydrogen diffused in 11.7 s under the same conditions. Calculate (i) the vapour density of the gas (ii) the molecular formula of the gas?

Solution:

(i). The two gases are of equal volumes,

t_x/t_H = √m_x /√m_H

45.2/11.7 = √(m_x/2)

 2(45.2/11.7)² = m_x

m_x = 2 x 14.92 = 29.84 (g)

r.m.m. = 2 x v.d.

v.d. = r.m.m./2

v.d. = 29.84/2 = 14.92

(ii). xCH₃ = 30

15x = 30

x = 2

Therefore, molecular formula = C₂H₆