Question

The specific conductance of a $0.1{\text{N}}$ ${\text{KCl}}$ solution is $0.012{\Omega ^{ - 1}}{\text{c}}{{\text{m}}^{ - 1}}$. The resistance of the cell containing the same solution is $55\Omega $. Then the cell constant is:
A. $0.918{\text{c}}{{\text{m}}^{ - 1}}$
B. $0.66{\text{c}}{{\text{m}}^{ - 1}}$
C. $1.142{\text{c}}{{\text{m}}^{ - 1}}$
D. $1.12{\text{c}}{{\text{m}}^{ - 1}}$

Answer 1

Hint: Specific conductance of an electrolyte solution is a measure of its ability to conduct electricity. The conductance of all electrolytes increases with temperature. Specific conductivity decreases on dilution. It depends upon the cell constant and the conductance.

Complete step by step answer:
Cell constant is the ratio of distance between two electrodes to the area of cross-section of the electrode. It is denoted by \[{{\text{G}}^ * }\].
From the definition, \[{{\text{G}}^ * } = \dfrac{{\text{l}}}{{\text{A}}}\], where \[{\text{l}}\] is the distance between two electrodes.
\[{\text{A}}\] is the area of cross-section of the electrode.
It is given that the specific conductance of ${\text{KCl}}$ solution, ${\text{ K= 0}}{\text{.012}}{\Omega ^{ - 1}}{\text{c}}{{\text{m}}^{ - 1}}$.
And resistance of the cell, ${\text{R = 55}}\Omega $
The resistance of a conductor is directly proportional to the length, ${\text{l}}$ and inversely proportional to the area of cross-section, ${\text{A}}$ of the conductor.
i.e. ${\text{R}}\alpha \dfrac{{\text{l}}}{{\text{A}}}$ which implies that ${\text{R}} = {{\rho }}\times\dfrac{{\text{l}}}{{\text{A}}}$, where $\rho $ is the proportionality constant and is called specific resistance. Conductance, ${\text{L}}$ is the reciprocal of resistance.
So $\dfrac{1}{\rho } = \dfrac{1}{{\text{R}}} \times \dfrac{{\text{l}}}{{\text{A}}}$
$\dfrac{1}{\rho } = {\text{L}} \times \dfrac{{\text{l}}}{{\text{A}}}$
Specific conductance of an electrolyte solution is defined as the conductivity of one cubic centimeter of the solution.
Thus ${\rm K} = {\text{L}} \times \dfrac{{\text{l}}}{{\text{A}}}$
Since $\dfrac{{\text{l}}}{{\text{A}}}$ is the cell constant denoted by ${{\text{G}}^ * }$
Therefore specific conductance can be represented as ${\rm K} = {\text{L}} \times {{\text{G}}^ * }$
Conductance can be calculated by ${\text{G}} = \dfrac{1}{{\text{R}}}$
Substituting the values, we get
${\text{G}} = \dfrac{1}{{55\Omega }} = 0.018{\Omega ^{ - 1}}$
Now we have to calculate the cell constant ${{\text{G}}^ * }$ from the values of conductance and specific conductance.
Substituting the values, we get $0.012{\Omega ^{ - 1}}.{\text{c}}{{\text{m}}^{ - 1}} = 0.018{\Omega ^{ - 1}} \times {{\text{G}}^ * }$
Cell constant, ${{\text{G}}^ * } = \dfrac{{0.012{\Omega ^{ - 1}}{\text{c}}{{\text{m}}^{ - 1}}}}{{0.018{\Omega ^{ - 1}}}} = 0.67{\text{c}}{{\text{m}}^{ - 1}}$
Hence the cell constant is $0.67{\text{c}}{{\text{m}}^{ - 1}}$

So, the correct answer is Option B .

Note:
Conductance is a measure of how much electrons are passed into the solution. It is due to the movement of ions. Cell constant is a property of a cell which is a constant value. Electrolytic conductance is mainly dependent on the concentration of ions and nature of electrolytes.