Question:
A conductivity cell containing \(0.01\) M KCl produces resistance \(600\) ohms at \(25^{\circ }C\). Find the cell constant if conductivity of \(0.01\) M KCl at the same temperature is \(0.0014\) \(\text{ohm}^{-1}\text{cm}^{-1}\).
A conductivity cell containing \(0.01\) M KCl produces resistance \(600\) ohms at \(25^{\circ }C\). Find the cell constant if conductivity of \(0.01\) M KCl at the same temperature is \(0.0014\) \(\text{ohm}^{-1}\text{cm}^{-1}\).
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Cell constant equals conductivity multiplied by resistance.
Updated On: Nov 12, 2026
- \(0.8\text{cm}^{-1}\)
- \(0.84\text{cm}^{-1}\)
- \(0.88\text{cm}^{-1}\)
- \(0.76\text{cm}^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Conductivity is \(\kappa=\frac{G^{*}}{R}\), where \(G^{*}\) is the cell constant, so \(G^{*}=\kappa R\).
Step 2: Substitute
\(\kappa=0.0014\ \Omega^{-1}\text{cm}^{-1}\) and \(R=600\ \Omega\).
Step 3: Calculation
\(G^{*}=0.0014\times600=0.84\ \text{cm}^{-1}\), option (B).
Final Answer:
The cell constant is 0.84 per cm. \[ \boxed{\text{(B) }0.84\text{cm}^{-1}} \]
Conductivity is \(\kappa=\frac{G^{*}}{R}\), where \(G^{*}\) is the cell constant, so \(G^{*}=\kappa R\).
Step 2: Substitute
\(\kappa=0.0014\ \Omega^{-1}\text{cm}^{-1}\) and \(R=600\ \Omega\).
Step 3: Calculation
\(G^{*}=0.0014\times600=0.84\ \text{cm}^{-1}\), option (B).
Final Answer:
The cell constant is 0.84 per cm. \[ \boxed{\text{(B) }0.84\text{cm}^{-1}} \]
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