The equivalent conductance of solution is …… .
[If cell constant is `1.25 cm^(-1)` and resistance of `N//10` solution is `2.5xx10^(3) Omega`].

The specific conductance of a N//10 KCI solution at 18^(@)C is 1.12 xx 10^(-2) mho cm^(-1) . The resistance of the solution contained in the cell is found to be 65 ohms. Calculate the cell constant.

The specific conductance of a N//10 KCI solution at 18^(@)C is 1.12 xx 10^(-2) mho cm^(-1) . The resistance of the solution contained in the cell is found to be 65 ohms. Calculate the cell constant.

The specific conductance of a N//10 KCI solution at 18^(@)C is 1.12 xx 10^(-2) mho cm^(-1) . The resistance of the solution contained in the cell is found to be 65 ohms. Calculate the cell constant.

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.

At 25^(@)C the equivalent conductance of butanoic acid at infinite dilution is 386.6 ohm^(-1)cm^(2)eq^(-1) . If the ionization consatant is 1.4xx10^(-5) , calculate equivalent conductance of 0.05 N butanoic acid solution at 25^(@)C(ohm^(-1)cm^(2)eq^(-1)) ?

At 25^(@)C the equivalent conductance of butanoic acid at infinite dilution is 386.6 ohm^(-1)cm^(2)eq^(-1) . If the ionization consatant is 1.4xx10^(-5) , calculate equivalent conductance of 0.05 N butanoic acid solution at 25^(@)C(ohm^(-1)cm^(2)eq^(-1)) ?

The resistance of 0.1 N solution of salt is found to be 2.5 xx 10^(3) Omega . The equivalent conductance of solution (cell constant = 1.15 cm^(-1) ) in Omega^(-1) cm^(2) eq^(-1) is

The resistance of a 1N solution of salt is 50 Omega . Calculate the equivalent conductance of the solution, if the two platinum electrodes in solution are 2.1 cm apart and each having an area of 4.2 cm^(2) .