Question

The resistance of 0.01 N solution of an electrolyte is 210 $\Omega$ at 298 K with a cell constant of 0.88 $c{m}^{-1}$. Calculate the conductivity and equivalent conductivity of the solution.

• A. $4.19\times {10}^{-3}oh{m}^{-1}c{m}^{-1},419.05Sc{m}^{2}e{q}^{-1}$
• B. $5.19\times {10}^{-3}oh{m}^{-1}c{m}^{-1},519.05Sc{m}^{2}e{q}^{-1}$
• C. $6.19\times {10}^{-3}oh{m}^{-1}c{m}^{-1},619.05Sc{m}^{2}e{q}^{-1}$
• D. $7.19\times {10}^{-3}oh{m}^{-1}c{m}^{-1},719.05Sc{m}^{2}e{q}^{-1}$

Answer 1

The correct option is A $4.19\times {10}^{-3}oh{m}^{-1}c{m}^{-1},419.05Sc{m}^{2}e{q}^{-1}$

First check what is given in the question.
$R=210ohmand\frac{1}{a}=0.88c{m}^{-1}Now.K=\frac{1}{R}\times \frac{1}{a}=\frac{1}{210}\times 0.88=4.19\times {10}^{-3}oh{m}^{-1}c{m}^{-1}orSc{m}^{-1}{\wedge }_{eq}=\frac{k\times 1000}{N}=\frac{4.19\times {10}^{-3}\times 1000}{0.01}=419.05Sc{m}^{2}e{q}^{-1}$