A gas X diffuse through porous partition at the rate of 3 cm3 per second. Under the same condition hydrogen…

Analyze the question: This is about Graham's law of diffusion. We compare the diffusion rate of gas X with that of hydrogen under the same conditions.

Concept: Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density):

Rate₁ / Rate₂ = √(M₂ / M₁)

Solution:

Rate of H₂ = 15 cm³/s, Rate of X = 3 cm³/s

Rate(H₂) / Rate(X) = √(M_X / M_H₂)

15 / 3 = √(M_X / M_H₂)

5 = √(M_X / M_H₂)

Squaring both sides:

25 = M_X / M_H₂

M_X = 25 × M_H₂

• If H₂ is taken as 2 (molecular hydrogen), M_X = 25 × 2 = 50
• If H is taken as 1, M_X = 25, which is not an option.

Answer: E. 50

Tip: In Graham's law questions, always use the molecular mass of the gas as it actually exists (H₂ = 2, O₂ = 32, etc.), not the atomic mass. Also remember that the slower gas is the heavier one. Here X diffuses 5 times slower, so its mass is 5² = 25 times greater than that of hydrogen.