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Question
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. Identify the gas
[Hint: Use Graham’s law of diffusion: R1/R2 = (M2/M1)1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
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Solution 1
Rate of diffusion of hydrogen, R1 = 28.7 cm3 s–1
Rate of diffusion of another gas, R2 = 7.2 cm3 s–1
According to Graham’s Law of diffusion, we have:
`R_1/R_2 = sqrt(M_2/M_1)`
Where,
M1 is the molecular mass of hydrogen = 2.020 g
M2 is the molecular mass of the unknown gas
`:. M_2 = M_1 (R_1/R_2)^2`
`= 2.02 (28.7/7.2)^2 = 32.09 g`
32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.
Solution 2
According to Graham’s law of diffusion of gases, the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.
If R1 and R2 be the rates of diffusion of two gases having molecular masses M1 and M2 respectively, then
`R_1/R_2 = sqrt(M_2/M_1)`
Now `R_1 = 28.7 cm^3 s^(-1), R_2 = 7.2 cm^3 s^(-1), M_1 = 2, M_2= ?`
`:. 28.7/7.2 = sqrt(M_2/2)`
`or M_2/2 = (28.7xx28.7)/(7.2xx7.2)`
or `M_2 = (2xx28.7xx28.7)/(7.2xx7.2) = 31.78 ~~ 32`
This molecular mass of oxygen.Therefore the second gas is oxygen.
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