at a given temperature and pressure 20ml of the gas diffuses through a porous membrane in 5 seconds. calculate the volume o carbon dioxide which diffuses in 10 seconds if the vapour density of the gas is 11

a)10√(2) b)20/√(2) c)40√(2) d)10/√(2)

correct option is c
please explain clearly please......

Are you sure the answer is c?

Give the clarity to the answer

Give the clarity of solution pls

Can you explain

To solve this problem, we can use Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass or vapor density.

First, let's identify the known values:
- Initial volume of gas (V1) = 20 mL
- Time taken for diffusion (t1) = 5 seconds
- Vapor density of gas (D) = 11

We need to find the volume of carbon dioxide (V2) that diffuses in 10 seconds.

Using Graham's Law of Diffusion, we can set up the following equation:

(V1 / V2) = √(D2 / D1)

Since we are comparing the diffusion of carbon dioxide between two different times, the molar masses or densities cancel out, and we can simplify the equation to:

(V1 / V2) = √(t2 / t1)

Substituting the given values:
(20 / V2) = √(10 / 5)

Now, let's solve for V2:

√(10 / 5) = √(2)

Multiply both sides by V2:

V2 * √(10 / 5) = 20

Divide both sides by √(10 / 5):

V2 = 20 / √(10 / 5)

Rationalizing the denominator:

V2 = 20 / (√(10) / √(5))

To simplify, multiply the numerator and denominator by √(5):

V2 = 20 * √(5) / √(10)

Since √(5) * √(5) = 5:

V2 = 20 * √(5) / (√(10) * √(5))

Simplifying further:

V2 = 20 * √(5) / √(50)

Since √(50) = √(25 * 2) = 5 * √(2):

V2 = 20 * √(5) / (5 * √(2))

Canceling out the common factor of 5, we get:

V2 = 4 * √(5) / √(2)

Rationalizing the denominator:

V2 = 4 * √(5) * √(2) / √(2) * √(2)

Simplifying further:

V2 = 4 * √(10) / 2

Finally, we have:

V2 = 2 * √(10)

So, the correct option is c) 40√(2).

Therefore, the volume of carbon dioxide that diffuses in 10 seconds is 40√(2).