Final answer:
Using Graham's law of effusion, the relative rate of diffusion between ¹H₂ and ²H₂ does not match the given options upon calculation, but the closest estimate for O₂ and O₃ would be approximately 0.816, closest to option b) 0.75.
Step-by-step explanation:
To calculate the relative rate of diffusion of gases, we use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (M).
The formula in terms of rates of two gases (rate1/rate2) is:
rate1/rate2 = sqrt(M2/M1)
In the case of ¹H₂ (molar mass 2.0 g/mol) and ²H₂ (molar mass 4.0 g/mol), we calculate as follows:
rate1/rate2 = sqrt(4.0 g/mol / 2.0 g/mol) = sqrt(2) = 1.414
However, since we are looking for the rate of ¹H₂ relative to ²H₂, we invert this to get:
rate¹H₂/rate²H₂ = 1 / sqrt(2) ≈ 0.707
This value does not match the provided options, suggesting a possible mistake in the question or the options given. Regarding the calculation for O₂ (molar mass 32.0 g/mol) compared with O₃ (molar mass 48.0 g/mol):
rateO₂/rateO₃ = sqrt(48.0 g/mol / 32.0 g/mol) = sqrt(1.5) = 1.225
Once again, inverting for the relative rate of O₂:
rateO₂/rateO₃ = 1 / sqrt(1.5) ≈ 0.816
This value is not precisely matched by the options; the closest option is b) 0.75.