(a) Define the following terms: (i) Limiting molar conductivity (ii) Fuel cell. (b) Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is 100 Omega. IF the resistance of the same cell when filled with 0.02 mol L^-1 KClsolution is 520 Omega, calculate the conductivity and molar conductivity of 0.02 mol L^-1 KCl solution. The conductivity of 0.1 mol L^-1 KCl solution is 1.29 times 10^-2 Omega^-1 cm^-1.

(a) Define the following terms: (i) Limiting molar conductivity (ii)

1.	(a) Define the following terms: (i) Limiting molar conductivity (ii) Fuel cell. (b) Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is 100 Omega. IF the resistance of the same cell when filled with 0.02 mol L^-1 KClsolution is 520 Omega, calculate the conductivity and molar conductivity of 0.02 mol L^-1 KCl solution. The conductivity of 0.1 mol L^-1 KCl solution is 1.29 times 10^-2 Omega^-1 cm^-1.
Answer» Solution :(a) (i) Limiting molar conductivity: When concentration approaches zero the conductivity is known as limiting molar CONDUCITIVITY. (ii) Fuel cell: A fuel cells which convert the energy of COMBUSTION of fuels to electrical energy.(b) Given, cell constant =`G^@` `= conductivity times Resistance ` `thereforeG^@=1.29S//m times 100 OMEGA` `=129 m^-1=1.29 cm^-1` Conductivity of `0.02 mol L^-1`, KCl solution = cell constant/resistance `K=G^*/R=(129 m^-1)/(520 Omega)=0.248 S m^-1` `=0.248 times 10^-2 Scm^-1` Concentration =`0.02 mol L^-1` `=1000 times 0.2 mol m^-3` `=20 mol m^-3` Molar conductivity =`A_m=K/C` `=(248 times 10^-3 Sm^-1)/(20 mol m^-3)` `=124 times 10^-4 S m^2 mol^-1` `=124 S cm^2 mol^-1`