60cm^3 of oxygen gas diffused through a porous partition in 50seconds. How long would it take 60cm^3 of sulphur (iv) oxide gas to diffuse through the same pot under the same conditions (H=1, CL=35.5, S=32,O=16)

To solve this problem, we need to 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.

First, we need to calculate the molar mass of both gases.

For oxygen (O2), the molar mass is 2*16 = 32 g/mol.

For sulphur (iv) oxide (SO2), the molar mass is 32 + 2*16 = 64 g/mol.

Next, we use Graham's law to find the time it would take for the sulphur (iv) oxide to diffuse. According to Graham's law, the rate of diffusion of oxygen (rO2)

To solve this problem, we need to 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.

First, we need to calculate the molar mass of both gases.

For oxygen (O2), the molar mass is 2*16 = 32 g/mol.

For sulfur (IV) oxide (SO2), the molar mass is 32 + 2*16 = 64 g/mol.

Next, we use Graham's law to find the time it would take for the sulfur (IV) oxide to diffuse. Let's denote the time for oxygen as t1 (which is 50 seconds), the time for sulfur (IV) oxide as t2 (which we're trying to find), the molar mass of oxygen as M1 (which is 32 g/mol), and the molar mass of sulfur (IV) oxide as M2 (which is 64 g/mol).

According to Graham's law, (t2/t1) = sqrt(M2/M1).

Substituting the known values, we get:

t2/50 = sqrt(64/32)

Solving for t2, we get:

t2 = 50 * sqrt(2) = 70.71 seconds.

So, it would take approximately 71 seconds for 60cm^3 of sulfur (IV) oxide gas to diffuse through the same pot under the same conditions.

To solve this problem, we need to 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.

First, we need to calculate the molar mass of both gases.

For oxygen (O2), the molar mass is 2*16 = 32 g/mol.

For sulfur (IV) oxide (SO2), the molar mass is 32 + 2*16 = 64 g/mol.

According to Graham's law, the rate of diffusion of oxygen (rO2) to the rate of diffusion of sulfur (IV) oxide (rSO2) is equal to the square root of the molar mass of sulfur (IV) oxide (M_SO2) to the molar mass of oxygen (M_O2).

So, rO2/rSO2 = sqrt(M_SO2/M_O2)

We know that rO2 = 60 cm^3/50 s = 1.2 cm^3/s. We can substitute this into the equation and solve for rSO2.

1.2 cm^3/s / rSO2 = sqrt(64 g/mol / 32 g/mol)

Solving for rSO2, we get rSO2 = 1.2 cm^3/s / sqrt(2) = 0.8485 cm^3/s.

Finally, to find the time it would take for 60 cm^3 of sulfur (IV) oxide to diffuse, we divide the volume by the rate of diffusion.

t = V / r = 60 cm^3 / 0.8485 cm^3/s = 70.7 seconds.

So, it would take approximately 70.7 seconds for 60 cm^3 of sulfur (IV) oxide gas to diffuse through the same pot under the same conditions.