[" A conductance cell when filled with "0.5MKCl" solution (conductivity "=6.67times10-3Omega^(-1)cm^(-1)" ) registers a resistance of "243Omega" .Its "],[" cell constant is "]

A conductance cell when filled with 0.5 M KCI solution (conductivity = 6.67 xx 10^(-3) Omega^(-1) cm^(-1) ) register a resistance of 243 Omega . Its cell constant is .

A conductivity cell when filled with 0.02 M KCl (conductivity = 0.002768 Omega^(-1) cm^(-1) ) has a resistance of 457.3 Omega . What will be the equivalent conductivity of 0.05 N CaCl_(2) solution if the same cell filled with this solution has a resistance of 2020?

The specific conductivity of 0.1 N KCl solution is 0.0129 Omega^(-1) cm^(-1) . The resistane of the solution in the cell is 100 Omega . The cell constant of the cell will be

A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (kappa=1.25 times 10^(-3)" "S" "cm^(-1)) in the cell and the measured resistance was 800Omega at 25^(@)C . The cell constant is,

The resistance of a conductivity cell when filled with 0.05M solution of an electrolyte x is 100 Omega at 40^(@)C . The same conductivity cell filled with 0.01M solution of electrolyte y has a resistance of 50 Omega . The conductivity of 0.05 M solution of electrolyte x is 1.0 xx 10^(-4) S cm^(-1) . Calculate (i) cell constant (ii) conductivity of 0.01 M solution (iii) molar conductivity of 0.01 M 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.

Calculating conductivity and molar conductivity: Resistance of a conductivity cell filled with 0.1 M KCl solution is 100 Omega . If the resistance of the same cell when filled with 0.02 M KCl solutions 520 Omega and the conductivity of 0.1 KCl solution is 1.29 S m , calculate the conductivity and moalr conductivity of 0.02 M KCl solution. Strategy : Calculate the cell constant with the help of 0.01 M KCl solution (both R and kappa are Known). Use the cell constant to determine the conductivity of 0.02 M KCl solution and finally find its molar conductivity using the molarity

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100Omega . is 1.29 S m^(-1) . Resistance of the same cell when filled with 0.2 M of the same solution is 520Omega . The The molar conductivity of 0.2 M solution of the electrolyte will be :

If specific conductivity of N// 50 KCl solution at 298 K is 0.002765 Omega^(-1) cm^(-1) and resistance of a cell contaning this solution is 100 Omega , calculate the cell constant.