(a) Define the following terms :
(i) Limiting molar conductivity,
(ii) Fuel cell
(b) Resistance of a conductivity cell filled with `0.1 molL^(-1)KCl` solution is 100 Ohm. If the resistance of the same cell when filled with `0.2 mol L^(-1)KCl` solution is 520 Ohm , calculate the conductivity and molar conductivity of `0.2 mol L^(-1) KCl` solution. The conductivity of `0.1 mol L^(-1) KCl` solution is ` 1.29xx10^(-1) Scm^(-1)` .

Explanation

(i) When concentration of an electrolyte approaches zero or the molar comductiveity of a solution at infinite dilution. Then its molar conductivity is known as limiting molar conductivity.
(ii) Fuel cells are the galvanic cells in which the energy of combustion of the fuels like hydrogen, methanol. etc is directly converted into electrical energy.
(b) Given that : Concentration of the KCl solution = `0.1 mol L^(-1)`
Resistance of cell filled with `0.1 mol L^(-1)`
KCl solution = 1000 ohm
Cell constant = G =conductivity `xx resistence 1.29 xx10^(-2)ohm^(-1) cm^(-1)xx100` ohm
`1.29 cm^(-1)`
Cell constant for a particular conductivity cell ia a consant.
Conductivity of `0..2 mol L-1 KCl` solution
`k= (G^(ast))/(R)= (1.29)/(520) = 2.48xx10^(-3)`
Now, Molar conductivity = `wedge_(m)`
`(kxx100)/(M)=(2.48xx10^(-3)xx1000)/(0.02) = 124 S cm^(2) mol^(-1) ` .