if 465 cm3 of sulphur iv oxide can diffuse through porous partition in 30seconds how long will an equal volume, 620cm3 of hydrogen sulphide take to diffuse through the same partition (h=1,s=32,o=16
Is 465 cc = 620 cc?
No
I don't know solve it
To determine how long an equal volume of hydrogen sulphide (H2S) will take to diffuse through the same partition, we need to calculate the molar masses of both substances and compare their rates of diffusion.
1. Calculate the molar mass of sulphur IV oxide (SO2):
- The molar mass of sulfur (S) is 32 g/mol
- The molar mass of oxygen (O) is 16 g/mol
- The molar mass of SO2 = 32 g/mol + (16 g/mol x 2) = 64 g/mol
2. Calculate the molar mass of hydrogen sulphide (H2S):
- The molar mass of hydrogen (H) is 1 g/mol
- The molar mass of sulfur (S) is 32 g/mol
- The molar mass of H2S = (1 g/mol x 2) + 32 g/mol = 34 g/mol
3. Calculate the ratio of the molar masses:
- SO2 : H2S = 64 g/mol : 34 g/mol
- Simplifying the ratio gives us 32 g/mol : 17 g/mol
4. Calculate the ratio of the volumes:
- 465 cm3 : 620 cm3 (given)
- Simplifying the ratio gives us 3 : 4
5. Calculate the rate of diffusion ratio:
- Divide the molar mass ratio by the volume ratio:
(32 g/mol : 17 g/mol) ÷ (3 : 4)
= (32 g/mol ÷ 3) : (17 g/mol ÷ 4)
= 10.67 g/mol : 4.25 g/mol
The rate of diffusion for hydrogen sulphide (H2S) is approximately 10.67 g/mol : 4.25 g/mol compared to sulphur IV oxide (SO2).
6. Calculate the time ratio of diffusion:
- Divide the rate of diffusion ratio by the molar mass ratio:
(10.67 g/mol : 4.25 g/mol) ÷ (32 g/mol : 17 g/mol)
= (10.67 g/mol ÷ 32 g/mol) : (4.25 g/mol ÷ 17 g/mol)
= 0.3334 : 1
This means that hydrogen sulphide (H2S) will diffuse approximately 0.3334 times the rate of sulphur IV oxide (SO2).
7. Determine the time taken for hydrogen sulphide to diffuse through the same partition:
- Multiply the time for sulphur IV oxide (30 seconds) by the time ratio:
30 seconds x 0.3334
≈ 9.997 seconds
Therefore, an equal volume of 620 cm3 of hydrogen sulphide (H2S) will take approximately 9.997 seconds to diffuse through the same partition.