Hydrogen diffused through a porous plug

1. Analyze the question

The question asks for the relative rate of diffusion of hydrogen gas ($H_2$) compared to oxygen gas ($O_2$) through a porous plug. This is a classic application of Graham’s Law of Diffusion.

2. Explain the concept

Graham’s Law of Diffusion states that the rate of diffusion ($r$) of a gas is inversely proportional to the square root of its molar mass ($M$):

When comparing two gases (Gas 1 and Gas 2), the formula is:

• Molar mass of Hydrogen ($H_2$): $1 \times 2 = 2 \text{ g/mol}$
• Molar mass of Oxygen ($O_2$): $16 \times 2 = 32 \text{ g/mol}$

3. Show the solution

Let $r_H$ be the rate of hydrogen and $r_O$ be the rate of oxygen.

Using the formula:

Substitute the molar masses:

This means $r_H = 4 \times r_O$. Therefore, hydrogen diffuses four times as fast as oxygen.

4. Identify the answer

Comparing our result to the given options, the correct choice is C.

5. Tip

When solving diffusion problems, always remember the mnemonic: "Lighter gases move faster." Since hydrogen is much lighter than oxygen, it must diffuse faster. If you ever get a result where the heavier gas is moving faster, check your calculation—you likely flipped the molar masses in the square root!