(a) State the relationship amongst cell constant of a cell , resistance of the solution in the cell and conductivity of the solution . How is molar conductivity of solute related to conductivity of its solution ?
(b) A voltanic cell is set up at `25^(@)C` with the following half-cells :
`Al|Al^(3+) (0.001M)` and `Ni|Ni^(2+) (0.50M)`
Calculate the cell voltage [`E_(Ni^(2+)|Ni)^(@) = -0.25V , E_(Al^(3+) |Al)^(@) = -1.66 V`]

Express the relationship amongst cell constant , resistance of the solution in the cell and conductivity of the solution . How is molar conductivity of a solute related to conductivity of its solution ?

Express the relation among cell constant , resistance of the solution in the cell and conductivity of the solution . How is molar conductivity of a solution related to its conductivity ?

Express the relation among cell constant , resistance of the solution in the cell and conductivity of the solution . How is molar conductivity of a solution related to its conductivity ?

State the relationship amongst cell constant of a cell resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of the solution is related to conductivity of its solution ? A voltaic cell is set up at 25^(@)C with the following half cells Al//Al^(3+) (0.001 M) and Ni//Ni^(2+)(0.50M) Calculate the cell voltage. [(E_(Ni^(2+)//Ni)^(@)= -0.25V","),(E_(Al^(3+)//Al)^(@)= -1.66V),(E_("Cell")^(@)=1.4602V)]

Express relation in cell constant, resistance of solution in the cell and 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.

Calculate the emf of the cell consisting the following half cells Al//Al^(3+)(0.001M), Ni//Ni^(2)(0.50M). Given that E_(Ni^(2+)//Ni)^(0)=-0.25V E_(Al^(3+)//Al)=-1.66V(log 8xx10^(-6)=-5.0969).