Specific conductivity of N/35 KCl at 298K is `0.002768 ohm^-1 cm^-1` and it has resistance of 520 ohm. A N/25 solution of a salt kept in the same cell was found to have a resistance of 300 ohm at 298K. Calculate the equivalent conductance of the solution.

A conductivity cell has its electrodes 1 cm apart and each electrode has area of cross-section 2 cm^(2) , when filled with M/50 solution of sodium acetate, the cell shows a resistance of 166.5 ohms. Calculate the molar conductance of sodium acetate solution at the givne concentration.

A potential difference of 20 V applied to the ends of a column of 0.1 M AgNO_(3) solution, 4 cm in diameter and 12 cm in length have a current of 20 A. calculate the molar conductance of the solution.

The conductivity of 0.35M NaCl solution at 298K is 0.025 S/cm. Calculate its molar conductivity .

The conductivity of 0.20 MKCl solution at 298 K is 0.0248 S cm^-1 . Calculate its molar conductivity.

The resistance of 0.1N solution of acetic acid 250 ohm. When measured in a cell constant 1.15 cm^-1 . The equilvalent conductance (in ohm^-1 cm^2 eq^-1 ) of 0.1 N acetic acid is

The resistance of 0.05 M NaOH solution in a cell having length 5 cm area of cross section 10 cm2 is 5.55 xx 10^3 ohm. Calculate its molar conductance.

The resistance of a conductivity cell with 0.1 M KCl solution is found to be 200 ohm at 298K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the same temperature is found to be 1100 ohm . Calculate: the molar conductivity of 0.02 M NaCl solution in S m^2 mol^-1 . Given: Conductivity of 0.1 M KCl solution at 298K = 1.29 S m^-1 .

The resistance of a conductivity cell with 0.1 M KCl solution is found to be 200 ohm at 298K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the same temperature is found to be 1100 ohm . Calculate: the cell constant of the cell in m^-1 . Given conductivity of 0.lM KCl is 1.29 Sm^(-1) ]

The resistance of 0.1N solution of formic acid is 200 ohm and cell constant is 2.0 cm^(-1) . The equivalent conductivity ( in Scm^(2) eq^(-1)) of 0.1N formic acid is :