Answer
Verified
37.2k+ 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 ${\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\left( {{\rm{gm}}\,{\rm{equiv}}} \right)^{ - {\rm{1}}}}$
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 ${\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\left( {{\rm{gm}}\,{\rm{equiv}}} \right)^{ - {\rm{1}}}}$ .
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 ${\rm{oh}}{{\rm{m}}^{ - {\rm{1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\left( {{\rm{gm}}\,{\rm{equiv}}} \right)^{ - {\rm{1}}}}$ .
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
Other Pages
The escape velocity from the earth is about 11 km second class 11 physics JEE_Main
The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main
The ratio of speed of sound in Hydrogen to that in class 11 physics JEE_MAIN
Lowering in vapour pressure is highest for A 02 m urea class 11 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Which of the following is a nonreducing sugar A Glucose class 12 chemistry JEE_Main