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Question: Answered & Verified by Expert
A conductivity cell containing $0.001 \mathrm{M} \mathrm{AgNO}_3$ solution develops resistance $6530 \mathrm{ohm}$ at $25^{\circ} \mathrm{C}$. Calculate the electrical conductivity of solution at same temperature if the cell constant is $0.653 \mathrm{~cm}^{-1}$.
ChemistryElectrochemistryMHT CETMHT CET 2023 (14 May Shift 2)
Options:
  • A $1.3 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$
  • B $1.5 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$
  • C $1.7 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$
  • D $1.0 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$
Solution:
2244 Upvotes Verified Answer
The correct answer is: $1.0 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$
$\begin{array}{ll} & \text { Cell constant }=\mathrm{k} \times \mathrm{R} \\ \therefore & 0.653=\mathrm{k} \times 6530 \\ \therefore & \mathrm{k}=\frac{0.653}{6530}=1 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}\end{array}$

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