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Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32u)
- a)28.2
- b)14.1
- c)7.09
- d)10.0
Correct answer is option 'B'. Can you explain this answer?
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To solve this question, 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.
Let's denote the rate of diffusion of a gas as V and the molar mass as M. According to Graham's law:
V1/V2 = √(M2/M1)
Given that 20 dm3 of SO2 diffuses through the porous partition in 60 seconds, we can write:
V1 = 20 dm3
t1 = 60 s
We need to find the volume of O2 that diffuses in 30 seconds. Therefore:
t2 = 30 s
Now, let's calculate the molar mass of SO2.
The atomic mass of sulfur is 32u, and the atomic mass of oxygen is 16u. Since SO2 has 1 sulfur atom and 2 oxygen atoms, its molar mass is:
M1 = (32 + 2 * 16) g/mol = 64 g/mol
Since we are comparing the diffusion rates of SO2 and O2, we can assume that the molar mass of O2 is M2 = 32 g/mol.
Now, let's substitute the given values into Graham's law and solve for V2:
V1/V2 = √(M2/M1)
20/V2 = √(32/64)
20/V2 = √(1/2)
20/V2 = 1/√2
V2 = 20 * √2
V2 ≈ 14.1 dm3
Therefore, the volume of O2 that diffuses under similar conditions in 30 seconds is approximately 14.1 dm3. This corresponds to option B.
Let's denote the rate of diffusion of a gas as V and the molar mass as M. According to Graham's law:
V1/V2 = √(M2/M1)
Given that 20 dm3 of SO2 diffuses through the porous partition in 60 seconds, we can write:
V1 = 20 dm3
t1 = 60 s
We need to find the volume of O2 that diffuses in 30 seconds. Therefore:
t2 = 30 s
Now, let's calculate the molar mass of SO2.
The atomic mass of sulfur is 32u, and the atomic mass of oxygen is 16u. Since SO2 has 1 sulfur atom and 2 oxygen atoms, its molar mass is:
M1 = (32 + 2 * 16) g/mol = 64 g/mol
Since we are comparing the diffusion rates of SO2 and O2, we can assume that the molar mass of O2 is M2 = 32 g/mol.
Now, let's substitute the given values into Graham's law and solve for V2:
V1/V2 = √(M2/M1)
20/V2 = √(32/64)
20/V2 = √(1/2)
20/V2 = 1/√2
V2 = 20 * √2
V2 ≈ 14.1 dm3
Therefore, the volume of O2 that diffuses under similar conditions in 30 seconds is approximately 14.1 dm3. This corresponds to option B.
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Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32u)a)28.2b)14.1c)7.09d)10.0Correct answer is option 'B'. Can you explain this answer?
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Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32u)a)28.2b)14.1c)7.09d)10.0Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32u)a)28.2b)14.1c)7.09d)10.0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32u)a)28.2b)14.1c)7.09d)10.0Correct answer is option 'B'. Can you explain this answer?.
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