0.1 M solution of an electrolyte `A^(+)B^(-)` placed in a conductivity cell wilth electrodes 4 cm apart and each with area of cross-section equal to 2 sq cm was found to have a resistance of `200 Omega`. The molar conductivity of the solution will be

conductivity cell has two electrodes 20 mm apart and of cross section area 1.8 cm^(2) . Find the cell constant.

1.0 N solution of a salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq cm in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity of the solution.

0.5 N solution of a salt placed between two platinum electrode 2.0cm apart and of area of cross-secton 4.0 sq. cm has a resistance of 25 ohms. Calculate the equivalent conductance of solution.

Resistance of a conductvity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Omega . The conductivity of this solution is 1.29 Sm^(-1) . Resistance of the same cell when filled with 0.02M of the same solution is 520 Omega . the molar conductivity of 0.02M solution of the electrolyte will be:

Calculate the conductivity of a wire of length 2 m, area of cross-sectionl 2 cm^(2) and resistance 10^(-4) Omega .