Question

The resistance of $N/10$ solution is found to be ${\rm{2}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ohm}}$. The equivalent conductance of the solution is (cell constant = ${\rm{1}}{\rm{.25}}\;{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}$)
A. ${\rm{2}}{\rm{.5}}\,{\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{equi}}{{\rm{v}}^{ - {\rm{1}}}}$
B. $K = \dfrac{1}{R} \times \dfrac{l}{a}$
C. ${\rm{2}}{\rm{.5}}\,{\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^ - }^{\rm{2}}{\rm{equi}}{{\rm{v}}^{ - {\rm{1}}}}$
D. ${\rm{5}}\,{\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{ - {\rm{2}}}}{\rm{equi}}{{\rm{v}}^{ - {\rm{1}}}}$

Answer 1

The equivalent conductance of the solution is defined as the conducting power of the ions that is produced when a one-gram equivalent of the electrolyte is dissolved in the solution. The unit of equivalent conductance is ${\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\left( {{\rm{gm}}\,{\rm{equiv}}} \right)^{ - {\rm{1}}}}$

Complete step by step answer:
Given, the resistance, $K = \dfrac{1}{R} \times \dfrac{l}{a} = 2.5 \times {10^3}$
Cell constant, $k = \dfrac{l}{a} = - 1.15{\rm{c}}{{\rm{m}}^{\rm{2}}}$
Normality, $N = 0.1N$
The relation between $R,k$and $K$ is,
$K = \dfrac{1}{R} \times \dfrac{l}{a}$
Here, $K$is the specific conductance
Now substituting the value of the specific conductance given we get,
$\begin{array}{c}K = \dfrac{1}{{2.5 \times {{10}^3}}} \times 1.15{\rm{c}}{{\rm{m}}^{ - 1}}\\\ = \dfrac{{1.15}}{{2.5 \times {{10}^3}}}{\rm{oh}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\end{array}$
We know the relation between ${\Lambda _{eq}}$and $K$ is,
${\Lambda _{eq}} = \dfrac{{K \times 1000}}{N}$
Now substituting the value of $N$and $K$ we get,
$\begin{array}{c}{\Lambda _{eq}} = \dfrac{{K \times 1000}}{{2.5 \times {{10}^3} \times 0.1}} \times 1000\\\ = 4.6\,oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}\end{array}$
This value can be approximately taken as, $\approx 5\,oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}$
Therefore, out of the given options, B is the correct option. Options A, C and D are incorrect.

Additional information:
Molar conductivity is the power of conducting the ions which are produced when one mole of an electrolyte is dissolved in a solution. Molar conductivity is expressed as,
$\Lambda = \dfrac{k}{M}$
Here, $M$is the molar concentration.
The value of Molar conductivity is expressed by lambda $\left( \lambda \right)$. The unit of Molar conductivity is expressed as ${\rm{Sc}}{{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - {\rm{1}}}}$.

Note:
We use Normality for measuring the concentration of a solution. It is denoted by “N”. Normality is also described as the number of mole or gram equivalents of the solute which is present in one litre of the solution. Units of Normality is expressed as, ${\rm{eq}}\,{{\rm{L}}^{ - 1}}$ or ${\rm{meq}}\,{{\rm{L}}^{ - 1}}$.
Specific conductivity is defined as the measure of the ability of a material to conduct electricity. Specific conductivity is represented by the symbol “К”.
Equivalent conductance is the measure of the net conductance of the ions which is produced by one-gram equivalent of the given substance. The unit of equivalent conductance is ${\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\left( {{\rm{gm}}\,{\rm{equiv}}} \right)^{ - {\rm{1}}}}$ .