**Gas Diffusion**

Diffusion is the process by which molecules move from an area of higher concentration to an area of lower concentration. The rate of diffusion depends on several factors, including the molecular weight of the gas. Heavier molecules tend to diffuse more slowly compared to lighter molecules. In this scenario, we are given that 50 ml of gas A diffuses through a membrane in the same time as 40 ml of gas B.

**Using Graham's Law of Diffusion**

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:

Rate A/Rate B = √(Molar Mass B/Molar Mass A)

where Rate A and Rate B represent the rates of diffusion of gases A and B, and Molar Mass A and Molar Mass B represent their respective molar masses.

**Determining the Molar Mass of Gas B**

We are given that the molar mass of gas A is 64. Let's assume that the rate of diffusion for gas A is 1 unit (it can be any arbitrary unit). Using Graham's Law of Diffusion, we can calculate the rate of diffusion for gas B:

Rate A/Rate B = √(Molar Mass B/Molar Mass A)

1/Rate B = √(Molar Mass B/64)

Squaring both sides of the equation, we get:

1/(Rate B)^2 = Molar Mass B/64

Molar Mass B = 64/(Rate B)^2

**Ratio of Diffusion Rates**

According to the given information, 50 ml of gas A diffuses in the same time as 40 ml of gas B. Therefore, the ratio of their diffusion rates can be calculated as:

Rate A/Rate B = Volume A/Volume B

Rate A/Rate B = 50 ml/40 ml

Rate A/Rate B = 5/4

**Calculating Molar Mass B**

Substituting the ratio of diffusion rates into the equation for Molar Mass B, we have:

1/(Rate B)^2 = Molar Mass B/64

1/(4/5)^2 = Molar Mass B/64

1/(16/25) = Molar Mass B/64

25/16 = Molar Mass B/64

Molar Mass B = (25/16) * 64

Molar Mass B = 100

Therefore, the molar mass of gas B is 100.

**Explanation**

In summary, using Graham's Law of Diffusion, we can determine the molar mass of gas B. By comparing the diffusion rates of gases A and B, we can establish a ratio. Substituting this ratio into the equation, we can solve for the molar mass of gas B, which is found to be 100.