JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The resistance of conductivity cell containing \(0.01\, M\, KCl\) solution at \(298\, K\) is \(1750 \,\Omega\). If the conductively of \(0.01\, M \,KCl\) solution at \(298\, K\) is \(0.152 \times 10^{-3} \,S \,cm ^{-1}\), then the cell constant of the conductivity cell is \(..........\,\times 10^{-3} \,cm ^{-1}\).
- A \(452\)
- B \(312\)
- C \(266\)
- D \(199\)
Answer & Solution
Correct Answer
(C) \(266\)
Step-by-step Solution
Detailed explanation
\(K =\frac{1}{ R } \times \text { cell cons } \tan t\) \(0.152 \times 10^{-3}=\frac{1}{1750}\) cell constant cell constant \(=266 \times 10^{-3}\)
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