Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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. ${\rm{5}}\,{\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{ - {\rm{2}}}}{\rm{equi}}{{\rm{v}}^{ - {\rm{1}}}}$

Last updated date: 18th Jul 2024
Total views: 62.4k
Views today: 1.62k
Verified
62.4k+ views
Hint: 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 ${\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\left( {{\rm{gm}}\,{\rm{equiv}}} \right)^{ - {\rm{1}}}}$

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.

$\Lambda = \dfrac{k}{M}$
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 ${\rm{Sc}}{{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - {\rm{1}}}}$.
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 ${\rm{meq}}\,{{\rm{L}}^{ - 1}}$.
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 ${\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\left( {{\rm{gm}}\,{\rm{equiv}}} \right)^{ - {\rm{1}}}}$ .