Helium effuses through a small opening at a rate of 1×10–9 mol s–1. An unknown gas at the same tempe

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

Answer: d