Question

(A) Define the following terms:
(i) Limiting molar conductivity
(ii) Fuel cell
(B) Resistance of a conductivity cell filled with $0.1mol{L}^{-1}KCl$ solution is $100\Omega$. If the resistance of the same cell when filled with $0.02mol{L}^{-1}KCl$ solution is $540\Omega$, calculate the conductivity and molar conductivity of $0.02mol{L}^{-1}KCl$ solution. The conductivity of $0.1mol{L}^{-1}KCl$ solution is $1.29\times {10}^{-2}{\Omega }^{-1}{cm}^{-1}$.

Answer 1

(A)

(i) When a concentration of an electrolyte approaches zero, 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 likes hydrogen, methanol. etc is directly convert 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 = 100 ohm

Cell constant = G* = conductivity $\times$ resistance

$1.29\times {10}^{-2}oh{m}^{-1}c{m}^{-1}\times 100ohm=1.29c{m}^{-1}=129m^{-1}$

Cell constant for a particular conductivity cell is constant.

Conductivity of 0.02 mol $L^{-1}$ KCl solution $=\frac{\text{Cell constant G*}}{\text{Resistance R}}=\frac{129m^{-1}}{520ohm}=0.248S{m}^{-1}$

Concentration = 0.02mol $l^{-1}$

$=1000\times 0.02mol{m}^{-3}=20mol{m}^{-33}$

Now,

Molar conductivity $={\Lambda }_m=\frac{k}{c}=\frac{248\times {10}^{-3}S{m}^{-1}}{20mol{m}^{-3}}=124\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Therefore the molar conductivity of $0.02mol{L}^{-1}KCl$ solution is $124\times {10}^{-4}S{m}^{2}mo{l}^{-1}$