Helium effuses through a small opening at a rate of 1×10^–9 mol s–1. An unknown gas at the same temperature and pressure is found to effuse through the same opening at a rate of 3.78×10–10 mol s–1. What is the molecular mass of the unknown gas? a. 4 g mol–1
b. 14 g mol–1
c. 18 g mol–1
d. 28 g mol–1
e. 40 g mol–1
Explanation: We are given the rates of effusion and should use Graham's rate of effusion formula:
R1/ R2 = √(M2/M1) where R1 is rate of effusion of gas 1, R2 is rate of effusion for gas 2. M1 is molar mass of gas 1 and M2 is molar mass of gas 2
Let's say that helium is gas 1.
R1 = 1×10^–9 mol s–1. M1 = 4g/mol (from the Periodic Table)
R2 = 3.78×10^–10 mol s–1 M2= ?
(1×10^–9 mol s–1)/ 3.78×10^–10 mol s–1 = √(M2/4g/mol)
M2= 27.99 g/mol