Answer to Question #169626 in Chemistry for Samantha Hernandez

Question

A container with a closed pinhole contains 1 mol of H₂(g) and 1 mol of He(g). When the pinhole is opened, H₂(g) starts to effuse at a rate of 0.48 mol/h. What is the rate of effusion of He(g) at that moment?

•  A. 1.21 mol/h
•  B. 0.34 mol/h
•  C. 0.68 mol/h
•  D. 0.24 mol/h
•  E. 0.95 mol/h

Answer 1

Solution:

Graham's law of effusion:

From Graham’s law, we have:

(rate of effusion of H₂) / (rate of effusion of He) = (M_He)^1/2 / (M_H2)^1/2

The molar mass of He is 4.0026 g/mol.

The molar mass of H₂ is 2.016 g/mol.

Hence,

(0.48) / (rate of effusion of He) = (4.0026)^1/2 / (2.016)^1/2

(0.48) / (rate of effusion of He) = (2.00065) / (1.41986)

rate of effusion of He = 0.48 × 1.41986 / 2.00065 = 0.34065 = 0.34

rate of effusion of He = 0.34 mol/h

Answer: B. 0.34 mol/h.