Question 13.12:
Rate of diffusion of another gas, R2 = 7.2 cm3 s–1
According to Graham’s Law of diffusion, we have:
R1 / R2 = (M2 / M1)1/2
Where,
M1 is the molecular mass of hydrogen = 2.020 g
M2 is the molecular mass of the unknown gas
∴ M2 = M1(R1 / R2)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.
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.]
[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.]
Solution:
Rate
of diffusion of hydrogen, R1
= 28.7 cm3
s–1Rate of diffusion of another gas, R2 = 7.2 cm3 s–1
According to Graham’s Law of diffusion, we have:
R1 / R2 = (M2 / M1)1/2
Where,
M1 is the molecular mass of hydrogen = 2.020 g
M2 is the molecular mass of the unknown gas
∴ M2 = M1(R1 / R2)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.
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