At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that

Analysis

This question tests your understanding of Graham’s Law of Diffusion, which relates the rate of diffusion of a gas to its molar mass (or density).

Concept: Graham’s Law of Diffusion

Graham’s Law 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, X and Y, the relationship is expressed as:

Where:

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

Step-by-Step Solution

1. Identify the given information: • Gas X diffuses twice as fast as Gas Y. This means $r_X = 2r_Y$, or $\frac{r_X}{r_Y} = 2$.

2. Substitute into the formula: $$2 = \sqrt{\frac{M_Y}{M_X}}$$

3. Solve for the relationship between the molar masses: • Square both sides to remove the square root: $$2^2 = \frac{M_Y}{M_X}$$ $$4 = \frac{M_Y}{M_X}$$ • Rearrange to solve for $M_Y$: $$M_Y = 4 \times M_X$$

4. Interpret the result: The molar mass of Gas Y is four times the molar mass of Gas X. Therefore, Gas Y is four times as heavy as Gas X.

Conclusion

Comparing our result to the options provided, we see that option B matches our calculation.

Exam Tip: Always remember the "Inverse Square" relationship. If a gas is faster, it must be lighter. Since the rate is squared in the formula, a speed difference of 2 becomes a mass difference of $2^2 = 4$. This quick mental math will save you time in the exam hall!