
At \[{25^ \circ }C\] specific conductance of a normal solution of \[KCl\] is \[0.002765\;mhos\;c{m^{ - 1}}\]. The resistance of the cell is \[400\;ohms\]. The cell constant is (\[c{m^{ - 1}}\]):
(a) 0.815
(b) 1.016
(c) 1.106
(d) 2.106
Answer
163.5k+ views
Hint: The cell constant is the measure of the theoretical distance between electrodes and the area of the cross-section. Whereas specific conductance is defined as the conductance of a substance to conduct electricity. And it is the reciprocal value of specific resistance.
Complete Step by Step Solution:
The value of the cell constant (\[K\]) can be calculated in the following way.
We know cell constant is \[K = \frac{l}{A}\], where l is the length in \[{\rm{cm}}\]and A is the area in \[{\rm{c}}{{\rm{m}}^2}\].
Therefore, the value of cell constant will be in \[{\rm{c}}{{\rm{m}}^{ - 1}}\] i.e, \[K = \frac{{cm}}{{c{m^2}}} = c{m^{ - 1}}\].
Also, \[R \propto \frac{l}{A}\] or \[R = \rho \frac{l}{A}\] where\[\rho \] is specific resistant.
The reciprocal value of specific resistance is known as specific conductance (\[k\]) or conductivity i.e., \[k = \frac{l}{\rho }\].
Hence mathematically, \[\frac{l}{R} = \frac{l}{\rho }.\frac{A}{l}\]
On rearranging,
\[\frac{l}{R} = k.\frac{A}{l}\]
\[\frac{l}{A} = k.R\]
\[K = k.R\]
or
(Cell constant \[ = \] Specific conductance\[ \times \] Resistant) … Eq. 1
Here, \[K = \] cell constant
\[k = \] Specific conductance
\[R = \] Resistant
Given in question,
The specific conductance (\[k\]) for \[KCl\] is \[0.002765\;{\Omega ^{ - 1}}c{m^{ - 1}}\] and resistant of the cell is \[400\Omega \]. By putting these values in equation 1, the value of the cell constant can easily be calculated.
\[K = 0.002765\;{\Omega ^{ - 1}}c{m^{ - 1}} \times 400\Omega \]
Hence, \[K = 1.106c{m^{ - 1}}\].
Hence, the above explanation denotes that options C is correct.
Note: The value of cell constant is depending upon the cross-section area of electrodes, the distance between the electrodes and on the nature of the electric fields. The value of the cell constant of a conductivity cell stays constant for a cell.
Complete Step by Step Solution:
The value of the cell constant (\[K\]) can be calculated in the following way.
We know cell constant is \[K = \frac{l}{A}\], where l is the length in \[{\rm{cm}}\]and A is the area in \[{\rm{c}}{{\rm{m}}^2}\].
Therefore, the value of cell constant will be in \[{\rm{c}}{{\rm{m}}^{ - 1}}\] i.e, \[K = \frac{{cm}}{{c{m^2}}} = c{m^{ - 1}}\].
Also, \[R \propto \frac{l}{A}\] or \[R = \rho \frac{l}{A}\] where\[\rho \] is specific resistant.
The reciprocal value of specific resistance is known as specific conductance (\[k\]) or conductivity i.e., \[k = \frac{l}{\rho }\].
Hence mathematically, \[\frac{l}{R} = \frac{l}{\rho }.\frac{A}{l}\]
On rearranging,
\[\frac{l}{R} = k.\frac{A}{l}\]
\[\frac{l}{A} = k.R\]
\[K = k.R\]
or
(Cell constant \[ = \] Specific conductance\[ \times \] Resistant) … Eq. 1
Here, \[K = \] cell constant
\[k = \] Specific conductance
\[R = \] Resistant
Given in question,
The specific conductance (\[k\]) for \[KCl\] is \[0.002765\;{\Omega ^{ - 1}}c{m^{ - 1}}\] and resistant of the cell is \[400\Omega \]. By putting these values in equation 1, the value of the cell constant can easily be calculated.
\[K = 0.002765\;{\Omega ^{ - 1}}c{m^{ - 1}} \times 400\Omega \]
Hence, \[K = 1.106c{m^{ - 1}}\].
Hence, the above explanation denotes that options C is correct.
Note: The value of cell constant is depending upon the cross-section area of electrodes, the distance between the electrodes and on the nature of the electric fields. The value of the cell constant of a conductivity cell stays constant for a cell.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
