What is the ratio of the rate of diffusion of carbon(IV) oxide to that of propane, under the same conditions? [H = 1; C= 12; 0= 16]

According to Graham’s law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. To compare two gases, we use: Rate₁ / Rate₂ = √(M₂ / M₁), where M₁ and M₂ are their molar masses.

First, let us find the molar masses. Carbon(IV) oxide is CO₂: molar mass = 12 + (2 × 16) = 12 + 32 = 44 g/mol. Propane is C₃H₈: molar mass = (3 × 12) + (8 × 1) = 36 + 8 = 44 g/mol.

Now we apply Graham’s law: Rate of CO₂ / Rate of C₃H₈ = √(44 / 44) = √1 = 1. So the ratio is 1:1.

Since both gases have exactly the same molar mass (44 g/mol), they diffuse at exactly the same rate under the same conditions of temperature and pressure. Whenever two gases have equal molar masses, their diffusion rates will be identical regardless of their chemical differences.