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A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1?
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A conductivity cell when filled with .01 M KCL has a resistance of 747...
To solve this question, we need to use the given information about the resistance of the conductivity cell filled with different solutions to calculate the conductivity and molar conductivity of the solution.
(i) Conductivity of the solution:
The resistance of the conductivity cell filled with the 0.01 M KCl solution is given as 747.5 Ω at 25ºC.
Using the formula:
Conductivity (σ) = 1 / Resistance (R)
We can calculate the conductivity of the 0.01 M KCl solution as:
σ1 = 1 / 747.5 Ω
(ii) Molar conductivity of the solution:
The molar conductivity (Λ) of a solution is the conductivity of a solution divided by the concentration of the solution.
The resistance of the conductivity cell filled with the 0.05 M CaCl solution is given as 876 Ω.
Using the formula:
Conductivity (σ) = Molar Conductivity (Λ) * Concentration (C)
We can rearrange the formula to solve for the molar conductivity:
Λ2 = σ2 / C2
Substituting the values:
Λ2 = 876 Ω / 0.05 M
To compare the two solutions, we can calculate the molar conductivity of the 0.01 M KCl solution using the same formula:
Λ1 = σ1 / C1
Substituting the values:
Λ1 = σ1 / 0.01 M
Now, we can calculate the molar conductivity of both solutions.
(iii) Comparison and explanation:
To compare the molar conductivities of the two solutions, we can take the ratio:
Λ2 / Λ1 = (σ2 / C2) / (σ1 / C1)
Substituting the values:
Λ2 / Λ1 = (876 Ω / 0.05 M) / (σ1 / 0.01 M)
Since we know the conductivity of the 0.01 M KCl solution is 0.14114 S cm-1, we can substitute that value into the equation:
Λ2 / Λ1 = (876 Ω / 0.05 M) / (0.14114 S cm-1 / 0.01 M)
Simplifying the equation will give us the ratio of the molar conductivities.
By following these steps and using the given information, we can calculate the conductivity and molar conductivity of the solution, and compare the molar conductivities of the two solutions.
(i) Conductivity of the solution:
The resistance of the conductivity cell filled with the 0.01 M KCl solution is given as 747.5 Ω at 25ºC.
Using the formula:
Conductivity (σ) = 1 / Resistance (R)
We can calculate the conductivity of the 0.01 M KCl solution as:
σ1 = 1 / 747.5 Ω
(ii) Molar conductivity of the solution:
The molar conductivity (Λ) of a solution is the conductivity of a solution divided by the concentration of the solution.
The resistance of the conductivity cell filled with the 0.05 M CaCl solution is given as 876 Ω.
Using the formula:
Conductivity (σ) = Molar Conductivity (Λ) * Concentration (C)
We can rearrange the formula to solve for the molar conductivity:
Λ2 = σ2 / C2
Substituting the values:
Λ2 = 876 Ω / 0.05 M
To compare the two solutions, we can calculate the molar conductivity of the 0.01 M KCl solution using the same formula:
Λ1 = σ1 / C1
Substituting the values:
Λ1 = σ1 / 0.01 M
Now, we can calculate the molar conductivity of both solutions.
(iii) Comparison and explanation:
To compare the molar conductivities of the two solutions, we can take the ratio:
Λ2 / Λ1 = (σ2 / C2) / (σ1 / C1)
Substituting the values:
Λ2 / Λ1 = (876 Ω / 0.05 M) / (σ1 / 0.01 M)
Since we know the conductivity of the 0.01 M KCl solution is 0.14114 S cm-1, we can substitute that value into the equation:
Λ2 / Λ1 = (876 Ω / 0.05 M) / (0.14114 S cm-1 / 0.01 M)
Simplifying the equation will give us the ratio of the molar conductivities.
By following these steps and using the given information, we can calculate the conductivity and molar conductivity of the solution, and compare the molar conductivities of the two solutions.
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A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1?
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A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1?.
A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1?.
Solutions for A conductivity cell when filled with .01 M KCL has a resistance of 747.5 25ºC. When the same cell was filled with an aq. Solution of .05 M CaCl solution the resistance was 876 Q. Calculate - (i) Conductivity of solution (ii) Molar conductivity of solution [Conductivity of 0.01M KCL= 0.14114 S cm-1? in English & in Hindi are available as part of our courses for Class 12.
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