
A KCl solution of conductivity 0.14 S m-1 shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops 1.03 Ω. The conductivity of the HCl solution is ………………. $\times {10^{-2}}~ S~{m^{-1}}$. (Round off to the Nearest Integer).
Answer
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Hint:The conductance of a cube with unit volume material is known as conductivity. The units of conductivity are \[S{m^{ - 1}}\]. The conductance is the property of the conductor which facilitates the flow of electricity through it. Conductance is the reciprocal of resistance. Its unit is mho.
Formula used:
$ \Rightarrow R = \rho \dfrac{l}{a}$
Complete 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.,
$ \Rightarrow R = \rho \dfrac{l}{a}$
$ \Rightarrow \dfrac{1}{\rho } = \sigma = \dfrac{1}{R} \times \dfrac{l}{a}$
Where ρ is called specific resistance depending upon the nature of the material and.
The inverse of the resistance is called conductance and similarly, the inverse of specific resistance is called specific conductance. $l$is the length of conductivity and a is the area of the cell.
For KCl solution,
\[ \Rightarrow \sigma \times R = \dfrac{l}{a}\]
$ \Rightarrow \dfrac{l}{a} = 4.19 \times 0.14 = 0.58ohm$
Now, For HCl solution since the same cell is filled their dimensions will not change so as the ratio of length to area will not change.
\[ \Rightarrow {\sigma _{new}} \times {R_{new}} = \dfrac{l}{a}\]
\[ \Rightarrow {\sigma _{new}} = \dfrac{l}{{a{R_{new}}}} = \dfrac{{0.58}}{{1.03}}\]
\[ \Rightarrow {\sigma _{new}} = 56 \times {10^{ - 2}}S{m^{ - 1}}\]
So, the correct answer is 56.
Additional information:
The resistance of a solution is determined by the Wheatstone bridge method using a meter bridge; the conductivity cell remains dipped in the test solution without disturbance. The current used is AC. The specific conductance of the 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.
Note: As the temperature increases, the conductivity of the solution increases. This is because as the temperature increases the interaction and the mobility of the ions in the solution increases. Also, upon dilution the number of ions in a weak electrolyte increases and the volume of solution increases. The conductivity should obviously decrease.
Formula used:
$ \Rightarrow R = \rho \dfrac{l}{a}$
Complete 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.,
$ \Rightarrow R = \rho \dfrac{l}{a}$
$ \Rightarrow \dfrac{1}{\rho } = \sigma = \dfrac{1}{R} \times \dfrac{l}{a}$
Where ρ is called specific resistance depending upon the nature of the material and.
The inverse of the resistance is called conductance and similarly, the inverse of specific resistance is called specific conductance. $l$is the length of conductivity and a is the area of the cell.
For KCl solution,
\[ \Rightarrow \sigma \times R = \dfrac{l}{a}\]
$ \Rightarrow \dfrac{l}{a} = 4.19 \times 0.14 = 0.58ohm$
Now, For HCl solution since the same cell is filled their dimensions will not change so as the ratio of length to area will not change.
\[ \Rightarrow {\sigma _{new}} \times {R_{new}} = \dfrac{l}{a}\]
\[ \Rightarrow {\sigma _{new}} = \dfrac{l}{{a{R_{new}}}} = \dfrac{{0.58}}{{1.03}}\]
\[ \Rightarrow {\sigma _{new}} = 56 \times {10^{ - 2}}S{m^{ - 1}}\]
So, the correct answer is 56.
Additional information:
The resistance of a solution is determined by the Wheatstone bridge method using a meter bridge; the conductivity cell remains dipped in the test solution without disturbance. The current used is AC. The specific conductance of the 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.
Note: As the temperature increases, the conductivity of the solution increases. This is because as the temperature increases the interaction and the mobility of the ions in the solution increases. Also, upon dilution the number of ions in a weak electrolyte increases and the volume of solution increases. The conductivity should obviously decrease.
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