Q.A conductivity cell containing 0.01M KCl gives at 25$^0$C the resistance of 604 ohms. The same cell containing 0.001M AgNO$_3$ gives resistance of 6530 ohms. Calculate the molar conductivity of 0.001M AgNO$_3$. [Conductivity of 0.01M KCl at 25 $^0$C is 0.00141 $\Omega^{-1}$ cm$^{-1}$]

The KCl measurement calibrates the cell (cell constant = $k_{KCl} \times R_{KCl}$); the same constant then converts the AgNO$_3$ resistance into its conductivity, and $\Lambda = 1000k/c$ finishes it. …

Step 1 (cell constant). The known KCl solution calibrates the cell:

Step 2 (conductivity of AgNO$_3$). With the same cell, $k = \dfrac{\text{cell constant}}{R} = \dfrac{0.852\ \mathrm{cm^{-1}}}{6530\ \Omega} = 1.3 \times 10^{-4}\ \Omega^{-1}\,\mathrm{cm^{-1}}$. …

Calibrate the cell with the KCl standard (cell constant = k x R), reuse the constant for the AgNO3 reading (k = cell constant/R), …

• Using the KCl concentration (0.01M) in the final step -- the molar conductivity asked for is of the 0.001M AgNO3 solution.
• Dividing the cell constant by the KCl resistance instead of the AgNO3 resistance in Step 2. …