Question

The resistance of ${\text{0}}{\text{.01N}}$NaCl solution is ${\text{200}}\Omega$ at ${\text{2}}{{\text{5}}^\circ }{\text{C}}$. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

Answer 1

To answer this question, we are provided with the values of resistance and normality and we have to find out the equivalent conductance which is defined as the conductivity strength of the ions which is produced when one mole of electrolyte is dissolved in the solution.
We will use resistance to find out the conductivity and then we can find out the equivalent conductance using conductivity and normality of the solution.
Formula Used-
${\Lambda _m}{\text{ = }}\dfrac{{{\text{k}} \times {\text{1000}}}}{{\text{C}}}$
where ${\Lambda _{\text{m}}}{\text{ = }}$molar conductivity
k=conductivity
C=concentration of a solution which can be molarity or normality.

Complete step by step answer:
Molarity conductivity can be defined as ${\Lambda _m}{\text{ = }}\dfrac{{{\text{k}} \times {\text{1000}}}}{{\text{C}}}$
Since in the question we are provided with values, i.e.,
${\text{R = 200}}\Omega$
Cell constant = $\dfrac{{\text{l}}}{{\text{A}}}$=1
${\text{N = 0}}{\text{.01N}}$
Since, here normality is given, we shall be replacing “C” with “N”
As we know that conductivity is inverse of resistance, so,
${\text{G = }}\dfrac{{\text{1}}}{{\text{R}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{200}}\Omega }}{\text{ = }}\dfrac{{\text{1}}}{{{\text{200}}}}{\text{S}}$
Now, we find the conductivity value:
${\text{k = }}\dfrac{{\text{G}}}{{{\text{l/A}}}}$ $= \dfrac{{\dfrac{1}{{200}}}}{1}$ $= \dfrac{1}{{200}}$
Now, to find out the equivalent conductance, we will use the equation:
${\Lambda _{eq}}{\text{ = }}\dfrac{{{{k \times 1000}}}}{{\text{N}}}$
Now, placing the values, we get:
${\Lambda _{eq}} = \dfrac{{\dfrac{1}{{200}} \times 1000}}{{0.01}} = 500{\text{Sc}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$

Note:
So as to clarify how we get the relation between conductivity and cell constant.
Conductance refers to how easily the current can flow through a conductor. Conductivity is hence defined as the inverse of the resistance.
First all of let us define resistance R as-
${\text{R = }}\dfrac{{\rho {\text{l}}}}{{\text{A}}}$
where R=resistance of the conductor
l= length of the conductor
A= cross-section area of the conductor
$\rho$= resistivity of the conductor
As conductance(G) of a material can be defined as the inverse of resistance so
${\text{G = }}\dfrac{{\text{1}}}{{\text{R}}}$ ${\text{ = }}\dfrac{{\text{1}}}{{\rho {\text{l/A}}}}$ ${\text{ = }}\dfrac{{\text{A}}}{{\rho {\text{l}}}}$ $= k \times$cell constant
where k=$\dfrac{{\text{1}}}{\rho }$ =conductivity of the material, cell constant= $\dfrac{{\text{l}}}{{\text{A}}}$