The specific conductivity of N/10 KCl solution at ${20^ \circ }C$ is $0.0120{\Omega ^{ - 1}}c{m^{ - 1}}$and the resistance of the cell containing this solution at ${20^ \circ }C$is $56\Omega $. The cell constant is.
Answer
600.6k+ views
Hint: The conductance is the property of the conductor (metallic as well as electrolytic) which facilitates the flow of electricity through it. Conductance is the reciprocal of resistance. It’s unit is mho.
Step by step answer: We know that the resistance (R) of a metallic conductor is directly proportional to its length and inversely proportional to its cross sectional area,i.e.,
$
R\alpha \dfrac{l}{a} \\
\Rightarrow R = \rho \dfrac{l}{a} \\
\Rightarrow \dfrac{1}{\rho } = \dfrac{1}{R}.\dfrac{l}{a} \\
$
Where $\rho $is a constant depending upon the nature of the material and is called specific resistance of the material. The reciprocal of the resistant is called conductance and similarly, the reciprocal of specific resistance is called specific conductance. As already mentioned we know that
$
\kappa = observed\; conductivity \times \dfrac{l}{a} \\
\Rightarrow \dfrac{l}{a} = x(cell\;constant) \\
\Rightarrow \kappa = \dfrac{1}{{observed\;resistance}} \times x \\
$
The resistance of a solution is determined by a wheatstone bridge method using a meter bridge; the conductivity cell remains dipped in the test solution. The current used is AC. The specific conductance of 0.1 N KCl solution is unknown. The resistance of 0.1 N KCl solution is first determined experimentally and thereby cell constant is calculated. The KCl solution is removed from the cell, it is washed with conductivity water and then filled with a test solution. The resistance of the test solution is measured and since specific conductivity of the solution is already known cell constant can be calculated. Therefore substituting the values in the above equations we get,
$
x = \kappa \times R \\
\Rightarrow x = 0.012 \times 56 \\
\Rightarrow x = 0.672 \\
$
Therefore the value of the cell constant is 0.672.
Note: Upon dilution the number of ions in the case of weak electrolyte increases and the volume of solution increases as well. The specific conductivity being the conductivity of 1 c.c. solution, should obviously decrease.
Step by step answer: We know that the resistance (R) of a metallic conductor is directly proportional to its length and inversely proportional to its cross sectional area,i.e.,
$
R\alpha \dfrac{l}{a} \\
\Rightarrow R = \rho \dfrac{l}{a} \\
\Rightarrow \dfrac{1}{\rho } = \dfrac{1}{R}.\dfrac{l}{a} \\
$
Where $\rho $is a constant depending upon the nature of the material and is called specific resistance of the material. The reciprocal of the resistant is called conductance and similarly, the reciprocal of specific resistance is called specific conductance. As already mentioned we know that
$
\kappa = observed\; conductivity \times \dfrac{l}{a} \\
\Rightarrow \dfrac{l}{a} = x(cell\;constant) \\
\Rightarrow \kappa = \dfrac{1}{{observed\;resistance}} \times x \\
$
The resistance of a solution is determined by a wheatstone bridge method using a meter bridge; the conductivity cell remains dipped in the test solution. The current used is AC. The specific conductance of 0.1 N KCl solution is unknown. The resistance of 0.1 N KCl solution is first determined experimentally and thereby cell constant is calculated. The KCl solution is removed from the cell, it is washed with conductivity water and then filled with a test solution. The resistance of the test solution is measured and since specific conductivity of the solution is already known cell constant can be calculated. Therefore substituting the values in the above equations we get,
$
x = \kappa \times R \\
\Rightarrow x = 0.012 \times 56 \\
\Rightarrow x = 0.672 \\
$
Therefore the value of the cell constant is 0.672.
Note: Upon dilution the number of ions in the case of weak electrolyte increases and the volume of solution increases as well. The specific conductivity being the conductivity of 1 c.c. solution, should obviously decrease.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

