If 465cm cube of sulphur (iv) oxide can diffuse through a porous pot in 30 seconds,how long will it take (I)an equal volume (II)620cm cube of hydrogen sulphide to diffuse through the same partition?(H=1,S=32,O=16)

Answer 1

From Graham's law of diffusion, we know that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass:

Rate = k / √(Molar mass)

Where:
Rate = rate of diffusion
k = constant
Molar mass = molar mass of the gas

Let's find the molar mass of both sulfur (IV) oxide (SO₂) and hydrogen sulfide (H₂S).

Molar mass of SO₂ = S + 2 * O = 32 + 2 * 16 = 64
Molar mass of H₂S = 2 * H + S = 2 * 1 + 32 = 34

Now let's find the rate of diffusion of SO₂:

SO₂ diffusion rate = k / √(64)

Next, let's find how long it takes for hydrogen sulfide to diffuse through the same partition for the two cases.

(I) Equal volume (465 cm³) of hydrogen sulfide:

Since the volume is the same, the rates of diffusion are equal:

SO₂ diffusion rate = H₂S diffusion rate
k / √(64) = k / √(34)

Removing k from both sides and solving for the time t₁:

t₁ = 30 * √(64) / √(34)

t₁ ≈ 30 * 8 / 5.83 ≈ 41.16 seconds

It will take approximately 41.16 seconds for an equal volume of hydrogen sulfide to diffuse through the partition.

(II) 620 cm³ of hydrogen sulfide:

To find the time it takes for 620 cm³ of hydrogen sulfide to diffuse, it's helpful to know the relationship between volume, rate, and time. Since rate = volume / time:

Time = Volume / Rate

Let's find the rate of diffusion for 620 cm³ of H₂S:

620 cm³ H₂S diffusion rate = (k / √(34)) * (620 / 465)

Now, let's find the time t₂ it would take for 620 cm³ of hydrogen sulfide to diffuse through the partition:

t₂ = 620 / (k / √(34)) * (620 / 465)

Using the SO₂ diffusion rate found earlier:

t₂ = 620 / (k / √(64)) * (30) * (620 / 465)

t₂ ≈ 620 / (8 / 5.83) * (30) * (1.333)

t₂ ≈ 81.94 seconds

It will take approximately 81.94 seconds for 620 cm³ of hydrogen sulfide to diffuse through the partition.