A gas X diffuses twice as fast as gas Y. if the relative molecular mass of X is 32, calculate the relative molecular mass of Y.

1. Analyze the question

We are given the relative rates of diffusion of two gases, X and Y, and the relative molecular mass of gas X. We need to find the relative molecular mass of gas Y. 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 relative molecular mass ($M$):

When comparing two gases, X and Y, the formula is:

Where:

• $r_X$ and $r_Y$ are the rates of diffusion of gases X and Y.
• $M_X$ and $M_Y$ are the relative molecular masses of gases X and Y.

3. Show the solution

Given:

• $r_X = 2 \times r_Y$ (Gas X diffuses twice as fast as gas Y)
• $M_X = 32$
• $M_Y = ?$

Step-by-step calculation:

1. Substitute the known values into the Graham's Law formula: $$\frac{2r_Y}{r_Y} = \sqrt{\frac{M_Y}{32}}$$

2. The $r_Y$ terms cancel out: $$2 = \sqrt{\frac{M_Y}{32}}$$

3. Square both sides to remove the square root: $$2^2 = \frac{M_Y}{32}$$ $$4 = \frac{M_Y}{32}$$

4. Solve for $M_Y$: $$M_Y = 4 \times 32$$ $$M_Y = 128$$

4. Identify the answer

The calculated relative molecular mass of gas Y is 128. This matches option A.

5. Exam Tip

When using Graham's Law, always remember the "Inverse Relationship" rule: the faster gas is always on top of the rate ratio, but its mass is on the bottom of the square root ratio. A common mistake is to flip the masses; if you get a mass that is smaller than expected for a slower gas, check your formula setup!