Question

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration ${\text{0}}{\text{.1 M}}$ is ${\text{ 100}}\Omega$ The conductivity of this solution is ${\text{1}}{\text{.29 S/m}}$. Resistance of the same cell when filled with ${\text{0}}{\text{.02 M}}$ is ${\text{520}}\Omega$. The molar conductivity of ${\text{0}}{\text{.02 M}}$ solution of the electrolyte will be.
1. ${\text{1}}{\text{24}} \times {\text{1}}{{\text{0}}^{ - 4}}{\text{ S}}{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}$
2. ${\text{1240 }} \times {\text{1}}{{\text{0}}^{ - 4}}{\text{ S}}{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}$
3. ${\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 4}}{\text{ S}}{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}$
4. ${\text{12}}{\text{.4}} \times {\text{1}}{{\text{0}}^{ - 4}}{\text{ S}}{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}$

Answer 1

Cell constant is the product of conductivity and resistance.
Use the following formula for the cell constant of the conductivity cell
${\text{Cell constant = Conductivity }} \times {\text{ resistance}}$.
Use the following formula to calculate molar conductivity.
${\text{Molar conductivity}} = \dfrac{{{\text{Conductivity}}}}{{{\text{1000 }} \times {\text{ molarity}}}}$

Complete answer:
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration ${\text{0}}{\text{.1 M}}$ is ${\text{ 100}}\Omega$ The conductivity of this solution is ${\text{1}}{\text{.29 S/m}}$. Calculate the cell constant.

{\text{Cell constant = }}1.29 \times {\text{ 100}} \\\ {\text{Cell constant = }}129{\text{ }}{{\text{m}}^{ - 1}} \\\\$$ The value of the cell constant of the conductivity cell is $$129{\text{ }}{{\text{m}}^{ - 1}}$$ Resistance of the same cell when filled with $${\text{0}}{\text{.02 M}}$$ is $${\text{520}}\Omega $$. Calculate the conductivity of the second solution $${\text{Conductivity}} = \dfrac{{{\text{Cell constant}}}}{{{\text{Resistance}}}} \\\ {\text{Conductivity}} = \dfrac{{129}}{{520}}{\text{ S }}{{\text{m}}^{ - 1}} \\\\$$ Hence, the conductivity of the second solution is $$\dfrac{{129}}{{520}}{\text{ S }}{{\text{m}}^{ - 1}}$$. Calculate the molar conductivity of second solution $${\text{Molar conductivity}} = \dfrac{{{\text{Conductivity}}}}{{{\text{1000 }} \times {\text{ molarity}}}} \\\ {\text{Molar conductivity}} = \dfrac{{\dfrac{{129}}{{520}}{\text{ S }}{{\text{m}}^{ - 1}}}}{{{\text{1000 }} \times {\text{ 0}}{\text{.02 M}}}} \\\ {\text{Molar conductivity}} = 124 \times {10^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} \\\\$$ Hence the molar conductivity of second solution is $$124 \times {10^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$$. _**Hence, the correct option is the option (1).**_ **Additional Information** In the equation for molar conductivity, the molarity is multiplied with 1000 to convert the unit of molarity from $${\text{mol d}}{{\text{m}}^{ - 3}}{\text{ to mol }}{{\text{m}}^{ - 3}}$$. This is done because in the numerator, the unit of conductivity is Siemens per metre. **Note:** The cell constant of the conductivity cell is independent of the concentration of the electrolyte. It depends on the dimensions of the electrode. For a given electrode, the cell constant is a constant. The cell constant is the ratio of distance between the electrodes to surface area of the electrodes.