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Question: The specific conductivity of N/10 KCl solution at \({20^ \circ }C\) is \(0.0120{\Omega ^{ - 1}}c{m^{...

The specific conductivity of N/10 KCl solution at 20∘C{20^ \circ }C is 0.0120Ω−1cm−10.0120{\Omega ^{ - 1}}c{m^{ - 1}}and the resistance of the cell containing this solution at 20∘C{20^ \circ }Cis 56Ω56\Omega . The cell constant is.

Explanation

Solution

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.,
Rαla ⇒R=ρla ⇒1ρ=1R.la  R\alpha \dfrac{l}{a} \\\ \Rightarrow R = \rho \dfrac{l}{a} \\\ \Rightarrow \dfrac{1}{\rho } = \dfrac{1}{R}.\dfrac{l}{a} \\\
Where ρ\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
κ=observed  conductivity×la ⇒la=x(cell  constant) ⇒κ=1observed  resistance×x  \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=κ×R ⇒x=0.012×56 ⇒x=0.672  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.