The resistance of `0. 01 N NaCl` solution at `25^@C` is `200 Omega` . Cell constant of conductivity cell is ` 1 cm^(-1)` . The equivalent conductnance is .

The resistance of 0.01 N solution at 25^(@)C is 200 ohm. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

The resistance of 0.01 N solution at 25^(@)C is 200 ohm. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

At 25^(@)C , the resistance of 0.01 N NaCl solution is 200 ohm . If cell constant of the conductivity cell is unity, then the equivalent conductance of the solution is:

(a) What are the two classifications of batteries ? What is the difference between them ? (b) The resistanc of 0.01 M NaCl solution at 25^(@)C is 200Omega. The cell constant of the conductivity cell is unity. Calculate the molar conductivity of the solution.

(a) What are the two classifications of batteries ? What is the difference between them ? (b) The resistanc of 0.01 M NaCl solution at 25^(@)C is 200Omega. The cell constant of the conductivity cell is unity. Calculate the molar conductivity of the solution.

The resistance of 0.01 M NaCl solution at 25°C is 200 ohm.The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution.

The resistance of 0.01 M NaCl solution at 25°C is 200 ohm.The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution.

The resistance of 0.1 N solution of acetic acid is 250 ohm , when measured in cell of cell constant 1.15 cm^(-1) . The equivalent conductance (in ohm^(-1) cm^(2) "equiv"^(-1) ) of 0.1 N acetic acid is :

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 :

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 :