Question

The resistance of 0.01 N solution of an electrolyte was found to be 210 ohms at 298 K using a conductivity cell of cell constant of $0.66{cm}^{-1}$. The value of equivalent conductance of solution is:

Answer 1

Let us consider the problem.

Let us consider the specific conductance is given by equation.

$k=\frac{K}{R}$

where,

K is the cell constant

R is the resistance

Implies that

$k=\frac{0.88}{220}$

implies that

$4\times {10}^{-3}on{m}^{-1}c{m}^{-1}$

Hence equivalent conductance of electrotype is

$\lambda =\frac{1000K}{C}$

=>

$\frac{1000\times 4\times {10}^{-3}}{0.1}$

implies that

$40oh{m}^{-1}c{m}^{2}e{q}^{-1}$