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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 m_{1} and m_{2} ; and rates of diffusion R_{1} and R_{2}:
R_{1}/R_{2} = √m_{2}/√m_{1} – (1)
R_{1}/R_{2} = √ρ_{2}v_{2} /√ρ_{1}v_{1} – (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 t_{1} and t_{2},
we have R_{1} = V/t_{1} and R_{2} = V/t_{2}
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,
R_{1}/R_{2} = √ρ_{2} /√ρ_{1} – (3)
R_{1}/R_{2} = t_{2}/t_{1} – (4)
Therefore,
R_{1}/R_{2} = t_{2}/t_{1} = √m_{2} /√m_{1} = (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.
Calculation Based on Graham's Law of Diffusion
30 cm^{3} of a gas, the empirical formula of which was CH_{3}, diffuses through a porous partition in 45.2 s. 30 cm^{3} 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). xCH_{3} = 30
15x = 30
x = 2
Therefore, molecular formula = C_{2}H_{6}
