Question

The resistance of 0.01 N $NaC$l solution is 200 $\Omega $ at ${25^o}C$. Cell constant of the conductivity cell is unity. Calculate the equivalent conductivity of the solution.

Answer 1

Hint: The formula to find the equivalent conductivity (${\Lambda _m}$) of a solution is: ${\Lambda _m} = \dfrac{{\kappa \times 1000}}{C}$. To find the equivalent conductivity, you must also know the formula for determining the conductivity ($\kappa $) of a solution. We are given that cell constant is unity and the quantity $\dfrac{l}{A}$ is known as cell constant.

Complete step by step solution:
Given that,
Concentration of $NaCl$ solution = 0.01 N
The resistance of $NaCl$ solution = 200 $\Omega $ at ${25^o}C$
Cell constant is unity. The quantity $\dfrac{l}{A}$ is called cell constant, where $l$ is the separation length of electrodes and $A$ is the cross-sectional area.
 Thus, $\dfrac{l}{A} = 1$
The equivalent conductivity of the solution can be determined by the equation:
${\Lambda _m} = \dfrac{{\kappa \times 1000}}{C}$
Here, ${\Lambda _m}$ is the equivalent conductivity, $\kappa $ is the conductivity and $C$ is the concentration of the solution.
We are given concentration of the solution but, to find the equivalent conductivity, we also need to find the conductivity of the solution.
Conductivity of the solution is given by the equation:
$\kappa = \dfrac{{{\text{Cell constant}}}}{{\text{R}}}$
Here, R is the resistance in ohm.
Therefore, substituting the given value of resistance i.e., 200 $\Omega $ and cell constant i.e., 1, we get:
$\kappa = \dfrac{1}{{200}}{\text{ }}S{\text{ }}c{m^{ - 1}}$
Now, substituting the given value of concentration of NaCl solution i.e., 0.01 N and conductivity (determined above), we get equivalent conductivity of NaCl solution as:
${\Lambda _m} = \dfrac{{\kappa \times 1000}}{C} = \dfrac{{1 \times 1000}}{{200 \times 0.01}}$
 $\therefore {\Lambda _m} = 500{\text{ }}Sc{m^2}mo{l^{ - 1}}$.

Thus, the equivalent conductivity of NaCl solution is 500 $Sc{m^2}mo{l^{ - 1}}$ and hence this is the required answer.

Note: Always take care of the units while finding the equivalent conductivity of the solution. In the equation, ${\Lambda _m} = \dfrac{{\kappa \times 1000}}{C}$, if $\kappa $ is expressed in $S{\text{ }}c{m^{ - 1}}$ and the concentration, $C$ in $mol{\text{ }}c{m^{ - 3}}$, then the units of ${\Lambda _m}$ are in $Sc{m^2}mo{l^{ - 1}}$. Cell constant which is equal to $\dfrac{l}{A}$ is usually denoted by the symbol, ${G^*}$.