140cm^3 of nitrogen gas diffuses through a membrane in 70seconds. How long will it take 200cm^3 of carbon(iv)oxide gas to diffuse through the same membrane under the same conditions of temperature and pressure

This problem can be solved using 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.

First, we need to find the molar mass of nitrogen and carbon(IV) oxide. The molar mass of nitrogen (N2) is approximately 28 g/mol and the molar mass of carbon(IV) oxide (CO2) is approximately 44 g/mol.

Next, we need to find the rate of diffusion of nitrogen. The rate of diffusion is the volume of gas that diffuses per unit time. In this case, the rate of diffusion of nitrogen is 140 cm^3 / 70 s = 2 cm^3/s.

According to Graham's law, the rate of diffusion of nitrogen divided by the rate of diffusion of carbon(IV) oxide is equal to the square root of the molar mass of carbon(IV) oxide divided by the molar mass of nitrogen.

Let's denote the rate of diffusion of carbon(IV) oxide as x. We can set up the following equation:

2 / x = sqrt(44 / 28)

Solving for x gives us x = 2 * sqrt(28 / 44) = 1.13 cm^3/s.

Finally, to find the time it takes for 200 cm^3 of carbon(IV) oxide to diffuse, we divide the volume by the rate of diffusion:

200 cm^3 / 1.13 cm^3/s = 177 seconds.

So, it will take approximately 177 seconds for 200 cm^3 of carbon(IV) oxide to diffuse through the same membrane under the same conditions of temperature and pressure.