Final answer:
To find the molecular weight of gas B when gas A has a molecular weight of 64 and they diffuse at rates of 50 mL and 40 mL respectively, we use Graham's law. The molecular weight of gas B is calculated to be 80.
Step-by-step explanation:
The question pertains to the diffusion of gases and the determination of the molecular weight of a gas based on the rate of diffusion. According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Using this principle, if two gases A and B diffuse through a membrane and it is given that 50 mL of gas A diffuses in the same time as 40 mL of gas B, we can set up a ratio of their rates of diffusion (rate of A to rate B) which will be equal to the inverse ratio of the square roots of their molar masses (sqrt(molar mass of B) to sqrt(molar mass of A)). Given the molar mass of A is 64, we can use this relationship to solve for the molar mass of B.
The formula derived from Graham's law is:
rate of diffusion of A / rate of diffusion of B = sqrt(molar mass of B) / sqrt(molar mass of A)
Substituting the given values:
50 mL / 40 mL = sqrt(molar mass of B) / sqrt(64)
We can solve for the molecular weight of B:
molar mass of B = (50/40)^2 × 64 = 80
Therefore, the molecular weight of gas B would be 80.