Question

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

• A. $419.05Cc{m}^{2}e{q}^{-1}$
• B. $219.05Cc{m}^{2}e{q}^{-1}$
• C. $719.05Cc{m}^{2}e{q}^{-1}$
• D. $919.05Cc{m}^{2}e{q}^{-1}$

Answer 1

The correct option is A $419.05Cc{m}^{2}e{q}^{-1}$

Value of resistance is given, cell constant is given
So from these we can find the specific conductance and from the specific conductance
We will find the equivalent conductivity
$R=210ohm.\frac{1}{a}=0.88c{m}^{-1}$
$k=\frac{1}{R}\times \frac{1}{a}=\frac{1}{210}\times 0.88=4.19\times {10}^{-3}oh{m}^{-1}c{m}^{-1}orSc{m}^{-1}$
${\wedge }_{eq}=\frac{k\times 1000}{N}=\frac{4.19\times {10}^{-3}\times 1000}{0.01}$
$=419.05Cc{m}^{2}e{q}^{-1}$