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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 s. The volume of O2 (in dm3) which diffuses in 30 s will be (atomic mass of sulphur = 32 u)
[JEE Main 2014]
- a)7.09
- b)14.1
- c)10.0
- d)28.2
Correct answer is option 'B'. Can you explain this answer?
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Sulphur dioxide and oxygen were allowed to diffuse through a porous pa...
Given:
Volume of SO2 diffused (V1) = 20 dm3
Time taken for SO2 to diffuse (t1) = 60 s
Atomic mass of sulphur (M) = 32 u
We need to find the volume of O2 (V2) that diffuses in 30 s.
Using Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Mathematically, we can express this as:
Rate of diffusion of gas 1 / Rate of diffusion of gas 2 = √(Molar mass of gas 2 / Molar mass of gas 1)
Let's consider the ratio of the rates of diffusion of SO2 and O2:
Rate of diffusion of SO2 / Rate of diffusion of O2 = √(Molar mass of O2 / Molar mass of SO2)
We can rewrite this equation as:
Rate of diffusion of O2 = (Rate of diffusion of SO2) × √(Molar mass of SO2 / Molar mass of O2)
Now, let's substitute the given values into the equation:
Rate of diffusion of O2 = (20 dm3 / 60 s) × √(32 u / Molar mass of O2)
To find the value of the molar mass of O2, we need to determine the ratio of the rates of diffusion of SO2 and O2.
Using the given information, we can determine this ratio:
Rate of diffusion of SO2 / Rate of diffusion of O2 = (20 dm3 / 60 s) / (V2 dm3 / 30 s)
Simplifying this equation:
Rate of diffusion of SO2 / Rate of diffusion of O2 = 20 / V2
Now, substitute this ratio into the previous equation:
(20 / V2) = √(32 u / Molar mass of O2)
Squaring both sides of the equation:
(20 / V2)^2 = 32 u / Molar mass of O2
Rearranging the equation:
Molar mass of O2 = (32 u) / ((20 / V2)^2)
Now, let's substitute the value of V2 = 30 s into the equation:
Molar mass of O2 = (32 u) / ((20 / 30)^2)
Molar mass of O2 = (32 u) / ((2/3)^2)
Molar mass of O2 = (32 u) / (4/9)
Molar mass of O2 = (32 u) × (9/4)
Molar mass of O2 = 72 u
Therefore, the molar mass of O2 is 72 u.
Substituting this value back into the equation for the rate of diffusion of O2:
Rate of diffusion of O2 = (20 dm3 / 60 s) × √(32 u / 72 u)
Rate of diffusion of O2 = (20 dm3 / 60 s) × √(4/9)
Rate of diffusion of O2 = (20 dm3 / 60 s) × (2/3)
Rate of diffusion of O2 = (1/3) × 20 dm3
Rate of diffusion of O2 = 20/3 dm3
Since the rate
Volume of SO2 diffused (V1) = 20 dm3
Time taken for SO2 to diffuse (t1) = 60 s
Atomic mass of sulphur (M) = 32 u
We need to find the volume of O2 (V2) that diffuses in 30 s.
Using Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Mathematically, we can express this as:
Rate of diffusion of gas 1 / Rate of diffusion of gas 2 = √(Molar mass of gas 2 / Molar mass of gas 1)
Let's consider the ratio of the rates of diffusion of SO2 and O2:
Rate of diffusion of SO2 / Rate of diffusion of O2 = √(Molar mass of O2 / Molar mass of SO2)
We can rewrite this equation as:
Rate of diffusion of O2 = (Rate of diffusion of SO2) × √(Molar mass of SO2 / Molar mass of O2)
Now, let's substitute the given values into the equation:
Rate of diffusion of O2 = (20 dm3 / 60 s) × √(32 u / Molar mass of O2)
To find the value of the molar mass of O2, we need to determine the ratio of the rates of diffusion of SO2 and O2.
Using the given information, we can determine this ratio:
Rate of diffusion of SO2 / Rate of diffusion of O2 = (20 dm3 / 60 s) / (V2 dm3 / 30 s)
Simplifying this equation:
Rate of diffusion of SO2 / Rate of diffusion of O2 = 20 / V2
Now, substitute this ratio into the previous equation:
(20 / V2) = √(32 u / Molar mass of O2)
Squaring both sides of the equation:
(20 / V2)^2 = 32 u / Molar mass of O2
Rearranging the equation:
Molar mass of O2 = (32 u) / ((20 / V2)^2)
Now, let's substitute the value of V2 = 30 s into the equation:
Molar mass of O2 = (32 u) / ((20 / 30)^2)
Molar mass of O2 = (32 u) / ((2/3)^2)
Molar mass of O2 = (32 u) / (4/9)
Molar mass of O2 = (32 u) × (9/4)
Molar mass of O2 = 72 u
Therefore, the molar mass of O2 is 72 u.
Substituting this value back into the equation for the rate of diffusion of O2:
Rate of diffusion of O2 = (20 dm3 / 60 s) × √(32 u / 72 u)
Rate of diffusion of O2 = (20 dm3 / 60 s) × √(4/9)
Rate of diffusion of O2 = (20 dm3 / 60 s) × (2/3)
Rate of diffusion of O2 = (1/3) × 20 dm3
Rate of diffusion of O2 = 20/3 dm3
Since the rate
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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 s. The volume of O2 (in dm3) which diffuses in 30 s will be (atomic mass of sulphur = 32 u)[JEE Main 2014]a)7.09b)14.1c)10.0d)28.2Correct 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 s. The volume of O2 (in dm3) which diffuses in 30 s will be (atomic mass of sulphur = 32 u)[JEE Main 2014]a)7.09b)14.1c)10.0d)28.2Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 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 s. The volume of O2 (in dm3) which diffuses in 30 s will be (atomic mass of sulphur = 32 u)[JEE Main 2014]a)7.09b)14.1c)10.0d)28.2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 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 s. The volume of O2 (in dm3) which diffuses in 30 s will be (atomic mass of sulphur = 32 u)[JEE Main 2014]a)7.09b)14.1c)10.0d)28.2Correct answer is option 'B'. Can you explain this answer?.
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