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If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).?
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If specific conductivity of N/50 KCl solution in a conductivity cell a...
**Given Data:**
- Specific conductivity of N/50 KCl solution (κ) = 0.002765 Ω^(-1) cm^(-1)
- Resistance of the conductivity cell (R) = 100 Ω
- Temperature (T) = 298 K
**To find:**
- Cell constant (G) in cm^(-1)
**Explanation:**
The specific conductivity (κ) of a solution is defined as the conductivity of the solution when 1 cm^3 of the solution is placed between two parallel plates with an area of 1 cm^2 and a distance of 1 cm.
The resistance of a conductivity cell (R) is the resistance offered by the solution in the cell to the flow of current.
The cell constant (G) is defined as the ratio of resistance (R) to the specific conductivity (κ) of the solution.
Mathematically, G = R/κ
**Substituting the given values into the formula:**
G = 100 Ω / 0.002765 Ω^(-1) cm^(-1)
**Calculating the value of G:**
G = 361.11 cm^(-1)
Therefore, the cell constant of the conductivity cell is 361.11 cm^(-1).
**Explanation of the Concept:**
The specific conductivity of a solution depends on the concentration of ions in the solution and their mobility. It is a measure of the ability of the solution to conduct electricity. The unit of specific conductivity is Ω^(-1) cm^(-1).
The resistance of a conductivity cell is the resistance offered by the solution in the cell to the flow of current. It is measured in ohms (Ω).
The cell constant is a characteristic property of a conductivity cell and is defined as the ratio of the resistance of the cell to the specific conductivity of the solution.
By knowing the resistance and specific conductivity of a solution, we can calculate the cell constant using the formula G = R/κ.
The cell constant is an important parameter as it is used to convert the measured conductivity of a solution into the actual conductivity. It helps in obtaining accurate and reliable results in conductivity measurements.
- Specific conductivity of N/50 KCl solution (κ) = 0.002765 Ω^(-1) cm^(-1)
- Resistance of the conductivity cell (R) = 100 Ω
- Temperature (T) = 298 K
**To find:**
- Cell constant (G) in cm^(-1)
**Explanation:**
The specific conductivity (κ) of a solution is defined as the conductivity of the solution when 1 cm^3 of the solution is placed between two parallel plates with an area of 1 cm^2 and a distance of 1 cm.
The resistance of a conductivity cell (R) is the resistance offered by the solution in the cell to the flow of current.
The cell constant (G) is defined as the ratio of resistance (R) to the specific conductivity (κ) of the solution.
Mathematically, G = R/κ
**Substituting the given values into the formula:**
G = 100 Ω / 0.002765 Ω^(-1) cm^(-1)
**Calculating the value of G:**
G = 361.11 cm^(-1)
Therefore, the cell constant of the conductivity cell is 361.11 cm^(-1).
**Explanation of the Concept:**
The specific conductivity of a solution depends on the concentration of ions in the solution and their mobility. It is a measure of the ability of the solution to conduct electricity. The unit of specific conductivity is Ω^(-1) cm^(-1).
The resistance of a conductivity cell is the resistance offered by the solution in the cell to the flow of current. It is measured in ohms (Ω).
The cell constant is a characteristic property of a conductivity cell and is defined as the ratio of the resistance of the cell to the specific conductivity of the solution.
By knowing the resistance and specific conductivity of a solution, we can calculate the cell constant using the formula G = R/κ.
The cell constant is an important parameter as it is used to convert the measured conductivity of a solution into the actual conductivity. It helps in obtaining accurate and reliable results in conductivity measurements.
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If specific conductivity of N/50 KCl solution in a conductivity cell a...
0.2765 per cm
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If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).?
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If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).?.
If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If specific conductivity of N/50 KCl solution in a conductivity cell at 298K is 0.002765 ohm^(-1) cm^(-1) providing a resistance of 100 ohms, the cell constant of cell is____ cm^(-1).?.
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