Question

The conductivity oh $0.1$ N NaOH solution is $0.022$ S cm${}^{-1}$. To this solution, an equal volume of $0.1$ N HCl solution is added which results into a decrease of the conductivity of solution to $0.0055$ S cm${}^{-1}$. The equivalent conductivity of NaCl solution is S cm${}^2$ equiv${}^{-1}$ is:

• A. $0.0055$
• B. $0.11$
• C. $110$
• D. $55$

Answer 1

The correct option is C $110$

$\begin{array}{c}\text{NaOH + HCl gives}2\text{free ions}\\ \text{So,}\Lambda =K\times \frac{1000}{N}\end{array}$

$\begin{array}{cc}\text{Where}& ,N=\text{equivalent conductivity}\\ N& =\text{normality}\end{array}$

$\begin{array}{c}\text{Putting the values and evaluating,}\\ N=2\times 0.0055\times \frac{1000}{0.1}=110{Scm}^2{eq}^{-1}\end{array}$

$\begin{array}{c}\text{So, the equivalent conductivity of NaCl solution}\\ \text{is}110{Scm}^2{eq}^{-1}\end{array}$