Question

Resistance of a conductivity cell filled with $0.1mol{{L}^{-1}}$KCl solution is $100\Omega$. If the resistance of the same cell when filled with $0.02mol{{L}^{-1}}$KCl solution is $520\Omega$, calculate the conductivity and molar conductivity of $0.02mol{{L}^{-1}}$ KCl solution. The conductivity of $0.1mol{{L}^{-1}}$ KCl is $1.29\times {{10}^{-2}}{{\Omega }^{-1}}c{{m}^{-1}}$.

Answer 1

In order to solve this question, we will be using the concept of conductivity and molar conductivity of a solution and also use its formulae to find the conductivity and the molar conductivity of the desired solution.

Formula used:
(A) Cell constant: $G=K\times R$
Where $K$ is the conductivity and $R$ is the resistance of the solution.
(B) Molar Conductivity: ${{\Lambda }_{m}}=\dfrac{K}{c}$
Where $C$ is the conductivity and $c$ is the concentration of the solution.

Complete step by step solution:
First of all, we will learn about a few terms mentioned in the question:
CONDUCTIVITY: The conductance of a material refers to the property of a material that allows ions to pass through it and thus conducts electricity. It is commonly defined as the reciprocal of the material's resistance. The conductance unit in SI is S. (Siemens). The ability of a substance to conduct electricity is measured by its specific conductivity (also known as conductivity).
MOLAR CONDUCTIVITY: The conductance of a volume of solution containing one mole of electrolyte held between two electrodes with the unit area of cross-section and distance of unit length is the molar conductivity of a solution at a given concentration. It is known as the ratio of specific conductivity to electrolyte concentration in general terms. It is represented by the symbol $\Lambda m$.
CELL CONSTANT: The cell constant of a given cell determines the ratio of the distance between the cell electrodes and the area of the electrodes. The distance between the electrodes (l) and their cross-sectional area determines the cell constant (A). It is calculated by measuring the resistance of a cell containing a conductivity solution with a defined conductivity.
First of all, we will write the data given to us,
Resistance of a conductivity cell filled with $0.1mol{{L}^{-1}}$ KCl solution ${{R}_{1}}=100\Omega$
Resistance of a conductivity cell filled with $0.02mol{{L}^{-1}}$ KCl solution ${{R}_{2}}=520\Omega$
The conductivity of $0.1mol{{L}^{-1}}$KCl solution ${{K}_{1}}=1.29\times {{10}^{-2}}{{\Omega }^{-1}}c{{m}^{-1}}$
We know that both the solutions are filled in the cell with the same cell constant. Therefore, using the formula for cell constant,
Cell constant: $G=K\times R$
As we are given the conductivity of $0.1mol{{L}^{-1}}$ KCl solution , using the value of its resistance and conductivity , we get
$\begin{aligned} & \Rightarrow G={{K}_{1}}\times {{R}_{1}} \\\ & \Rightarrow G=\left( 1.29\times {{10}^{-2}}{{\Omega }^{-1}}c{{m}^{-1}} \right)\left( 100\Omega \right) \\\ & \Rightarrow G=1.29c{{m}^{-1}} \\\ \end{aligned}$
As the cell constant is same for both the solutions, we can find the conductivity of $0.02mol{{L}^{-1}}$ KCl solution by writing
${{K}_{2}}=\dfrac{G}{{{R}_{2}}}$
Placing the values in above equation,
$\begin{aligned} & \Rightarrow {{K}_{2}}=\dfrac{1.29c{{m}^{-1}}}{520\Omega } \\\ & \Rightarrow {{K}_{2}}=2.48\times {{10}^{-3}}{{\Omega }^{-1}}c{{m}^{-1}} \\\ \end{aligned}$
Now, we will use the formula for molar conductivity and find the molar conductivity of $0.02mol{{L}^{-1}}$ KCl solution. Therefore,
Molar Conductivity: ${{\Lambda }_{m}}=\dfrac{{{K}_{2}}}{c}$
Where $c$ is the concentration of the solution.
$\left[ \begin{aligned} & c=0.02mol{{L}^{-1}} \\\ & \Rightarrow c=0.02\times {{10}^{-3}}molc{{m}^{-3}} \\\ \end{aligned} \right]$
$\begin{aligned} & \Rightarrow {{\Lambda }_{m2}}=\dfrac{2.48\times {{10}^{-3}}{{\Omega }^{-1}}c{{m}^{-1}}}{0.02\times {{10}^{-3}}molc{{m}^{-3}}} \\\ & \Rightarrow {{\Lambda }_{m2}}=124{{\Omega }^{-1}}mo{{l}^{-1}}c{{m}^{2}} \\\ \end{aligned}$
Therefore, we found the conductivity and the molar conductivity of $0.02mol{{L}^{-1}}$ KCl solution to be ${{K}_{2}}=2.48\times {{10}^{-3}}{{\Omega }^{-1}}c{{m}^{-1}}$ and ${{\Lambda }_{m2}}=124{{\Omega }^{-1}}mo{{l}^{-1}}c{{m}^{2}}$ respectively.

Note:
It is very important to keep in mind the units of each term when doing the calculation. There are many metric conversions to be done. Also in writing the unit of conductivity, sometimes it may also be given in terms of Siemens ($S$) which is the inverse of ohm ($\Omega$). As in the molar conductivity we calculated, it can also be written as ${{\Lambda }_{m2}}=124Smo{{l}^{-1}}c{{m}^{2}}$instead of ${{\Lambda }_{m2}}=124{{\Omega }^{-1}}mo{{l}^{-1}}c{{m}^{2}}$.