The rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Let the molecular mass of gas A be M and the molar mass of CO2 is 44 g/mol.
Rate of effusion of gas A = 40 ml / 140 s = 0.286 ml/s
Rate of effusion of CO2 = 40 ml / 110 s = 0.364 ml/s
According to Graham's Law of Effusion:
(rate of effusion of A) / (rate of effusion of CO2) = sqrt(Molar mass of CO2 / Molar mass of A)
0.286 / 0.364 = sqrt(44 / M)
Solving for M:
0.78636 = sqrt(44 / M)
0.78636^2 = 44 / M
0.61892 = 44 / M
M = 44 / 0.61892
M = 71.05 g/mol
Therefore, the molecular mass of gas A is approximately 71.05 g/mol.
Since the molar mass of gas A is closer to the molar mass of N2 (28.02 g/mol) or O2 (32 g/mol) than to CO2, we can predict that gas A is likely nitrogen (N2) or oxygen (O2).
40 ml of gas A effuse through a pin-hole in 140 seconds. The same volume of CO2 under
identical conditions effuses in 110 seconds. Calculate the molecular mass of gas A and predict its
identity.
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