Graham’s law application to perfume diffusion:\nPerfume X diffuses twice as fast as perfume Y. If the vapor density of gas X is 2 (relative to hydrogen), estimate the molecular weight of Y.

Introduction / Context:
Graham’s law relates diffusion rates of gases to the square roots of their molecular weights. This principle enables quick identification or cross-checking of gas identities in labs and process troubleshooting.

Given Data / Assumptions:

• Rate_X = 2 · Rate_Y (X diffuses twice as fast as Y).
• Vapor density of X (relative to H2) = 2 → molecular weight M_X = 2 × 2 = 4.
• Ideal behavior; same conditions of T and P.

Concept / Approach:
Graham’s law: Rate ∝ 1/√M. Thus Rate_X / Rate_Y = √(M_Y / M_X). Rearranging provides M_Y directly from the rate ratio and M_X.

Step-by-Step Solution:
Write 2 = √(M_Y / M_X).Square: 4 = M_Y / M_X → M_Y = 4 · M_X.Given M_X = 4 → M_Y = 4 × 4 = 16.

Verification / Alternative check:
If Y has M = 16, then Rate_X / Rate_Y = √(16/4) = √4 = 2, which matches the statement.

Why Other Options Are Wrong:
2, 4, and 8 correspond to smaller molecular weights that would not yield a rate ratio of 2; 32 would give Rate_X / Rate_Y = √(32/4) ≈ 2.828, too large.

Common Pitfalls:
Confusing vapor density with molecular weight; forgetting to square the rate ratio when solving for M_Y.

Final Answer:
16