A conductivity cell has a cell constant of `0.5 cm^(-1)`. This cell when filled with 0.01 M NaCl solution has a resistance of 384 ohms at `25^(@)C`. Calculate the equivalent conductance of the given solution.

A conductivity cell has a cell constant of "0.5 cm"^(-1) . This cell when filled with 0.01 M NaCl solution has a resistance of "384 ohm at "25^(@)C . Calculate the equivalent conductance of the given solution.

A 0.1N solution of NaCl has a specific conductance of 0.00112 ohm^-1 cm^-1 . Calculate the equivalent conductance of the solution.

The cell constant of a conductivity cell is "0.228 cm"^(-1) . If the cell is filled with "0.005 (N) K_(2)SO_(4) solution, the resistance recorded is 326 ohm at 25^(@)C . What will be (i) the specific conductance and (ii) the equivalent conductivity 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.

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 conductivity cell when filled with 0.01 M KCl has a resistance of 745 Omega at 25^(@) C . When the same cell was filled with an aqueous solution of 0.005 M CaCl_(2) solution the resistance was 874 Omega . Calculate (i) Conductivity of solution (ii) Molar conductivity of solution . [Conductivity of 0.01 M KCl = 0.141 S m^(-1) ]

Resistance of a conductivity cell filled with 0.1 M KCl is 100 ohm. If the resistence of the same cell when filled with 0.02 M KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCl solution. Conductivity of 0.1 KCl solution is 1.29xx10^(-2)" ohm "^(-1)cm^(-1) .