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.

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.

A. 128

B. 8

C. 16

D. 64

This is a Graham’s Law of Diffusion problem:

Rate of diffusion <>< 1/Molar mass

Rate of X/Rate of Y = MY/MX

Given:

Rate of X : Rate of Y = 2 : 1

* (MX = 32)

* (MY = ?)

**Step 1: Apply Graham’s Law

2 = sqrt{{MY}/{32}}

Step 2: Square both sides

2^2 = {MY}/{32} 

4 = {MY}/{32}

MY = 4*32 = 128

✅ Answer: A. 128

PQN: 10