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A sample of water from a large swimming pool has a resistance of 10000 ohm at 25ºC when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25ºC. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 Ω.
[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1 and molar conductivity of KCl at 0.02 M is 200 Ω-1cm2 mol-1.]
Q.
Volume (in Litres) of water in the pool is:
- a)1.25 × 105
- b)1250
- c)12500
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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A sample of water from a large swimming pool has a resistance of 10000...
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A sample of water from a large swimming pool has a resistance of 10000...
To solve this problem, we need to use the concept of conductance and resistance, as well as the molar conductance of the given solutions. Let's break down the problem into steps:
Step 1: Calculate the conductance of the water sample.
Given that the resistance of the water sample is 10000 ohm at 25°C, we can calculate the conductance using the formula:
Conductance (G) = 1 / Resistance (R)
Conductance = 1 / 10000 = 0.0001 S
Step 2: Calculate the molar conductance of the KCl solution.
Given that the resistance of the KCl solution is 100 ohm at 25°C, we can calculate the conductance using the formula:
Conductance (G) = 1 / Resistance (R)
Conductance = 1 / 100 = 0.01 S
Now, to calculate the molar conductance, we need to divide the conductance by the concentration of the KCl solution:
Molar Conductance = Conductance / Concentration
Molar Conductance = 0.01 / 0.02 = 0.5 S.m^2/mol
Step 3: Calculate the molar conductivity of the KCl solution.
The molar conductivity is given as 200 Ω^-1cm^2mol^-1 for the KCl solution at a concentration of 0.02 M.
Step 4: Calculate the molar conductivity of the NaCl solution.
We are given that 585 g of NaCl were dissolved in the pool, and the sample of the solution has a resistance of 8000 ohm at 25°C. Using the given molar conductance of NaCl at that concentration (125 Ω^-1cm^2mol^-1), we can calculate the molar conductivity using the formula:
Molar Conductivity = Molar Conductance / Concentration
Molar Conductivity = 125 / (mass of NaCl in grams / molar mass of NaCl)
Molar Conductivity = 125 / (585 / 58.5) = 1250 S.m^2/mol
Step 5: Calculate the molar concentration of the NaCl solution.
To calculate the molar concentration, we need to divide the molar conductivity by the molar conductance:
Molar Concentration = Molar Conductivity / Molar Conductance
Molar Concentration = 1250 / 0.5 = 2500 M
Step 6: Calculate the volume of water in the pool.
Given that the concentration of the NaCl solution is 2500 M, we can calculate the volume of water using the formula:
Volume (in Litres) = Mass of NaCl / (Molar Mass of NaCl * Molar Concentration)
Volume = 585 / (58.5 * 2500) = 0.004 L = 1.25 * 10^5 L
Therefore, the volume of water in the pool is 1.25 * 10^5 L, which corresponds to option A.
Step 1: Calculate the conductance of the water sample.
Given that the resistance of the water sample is 10000 ohm at 25°C, we can calculate the conductance using the formula:
Conductance (G) = 1 / Resistance (R)
Conductance = 1 / 10000 = 0.0001 S
Step 2: Calculate the molar conductance of the KCl solution.
Given that the resistance of the KCl solution is 100 ohm at 25°C, we can calculate the conductance using the formula:
Conductance (G) = 1 / Resistance (R)
Conductance = 1 / 100 = 0.01 S
Now, to calculate the molar conductance, we need to divide the conductance by the concentration of the KCl solution:
Molar Conductance = Conductance / Concentration
Molar Conductance = 0.01 / 0.02 = 0.5 S.m^2/mol
Step 3: Calculate the molar conductivity of the KCl solution.
The molar conductivity is given as 200 Ω^-1cm^2mol^-1 for the KCl solution at a concentration of 0.02 M.
Step 4: Calculate the molar conductivity of the NaCl solution.
We are given that 585 g of NaCl were dissolved in the pool, and the sample of the solution has a resistance of 8000 ohm at 25°C. Using the given molar conductance of NaCl at that concentration (125 Ω^-1cm^2mol^-1), we can calculate the molar conductivity using the formula:
Molar Conductivity = Molar Conductance / Concentration
Molar Conductivity = 125 / (mass of NaCl in grams / molar mass of NaCl)
Molar Conductivity = 125 / (585 / 58.5) = 1250 S.m^2/mol
Step 5: Calculate the molar concentration of the NaCl solution.
To calculate the molar concentration, we need to divide the molar conductivity by the molar conductance:
Molar Concentration = Molar Conductivity / Molar Conductance
Molar Concentration = 1250 / 0.5 = 2500 M
Step 6: Calculate the volume of water in the pool.
Given that the concentration of the NaCl solution is 2500 M, we can calculate the volume of water using the formula:
Volume (in Litres) = Mass of NaCl / (Molar Mass of NaCl * Molar Concentration)
Volume = 585 / (58.5 * 2500) = 0.004 L = 1.25 * 10^5 L
Therefore, the volume of water in the pool is 1.25 * 10^5 L, which corresponds to option A.
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A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. 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 A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. 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 A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A sample of water from a large swimming pool has a resistance of 10000 ohm at 25C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 100 ohm at 25C. 585 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of 8000 .[Given : Molar conductance of NaCl at that concentration is 125 ohm-1cm2mol-1and molar conductivity of KCl at 0.02 M is 200 -1cm2mol-1.]Q. Volume (in Litres) of water in the pool is:a)1.25 105b)1250c)12500d)None of theseCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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