(C) The Kohlrausch law for strong electrolytes like $NaCl$ is given by the equation: $\lambda_c = \lambda_\infty - b\sqrt{C}$.Rearranging this equatio

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$\lambda_\infty = \lambda_c - b\sqrt{C}$

(C) The Kohlrausch law for strong electrolytes like $NaCl$ is given by the equation: $\lambda_c = \lambda_\infty - b\sqrt{C}$.Rearranging this equation,we get: $\lambda_\infty = \lambda_c + b\sqrt{C}$.Multiplying the original equation by $\sqrt{C}$,we get: $\sqrt{C} \lambda_c = \sqrt{C} \lambda_\infty - bC$,which can be written as $\sqrt{C} \lambda_\infty = \sqrt{C} \lambda_c + bC$.Comparing these with the given options,the relation $\lambda_\infty = \lambda_c - b\sqrt{C}$ is incorrect because it contradicts the standard Kohlrausch equation.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Medium

Which of the following is not a correct relationship between $\lambda_c$ and $\lambda_\infty$ for $NaCl$? Here,$\lambda_c$ and $\lambda_\infty$ are the equivalent conductance at concentration $C$ and at infinite dilution,respectively.

A
$\lambda_\infty = \lambda_c + b\sqrt{C}$
B
$\sqrt{C} \lambda_\infty = \sqrt{C} \lambda_c + bC$
C
$\lambda_\infty = \lambda_c - b\sqrt{C}$
D
$\lambda_c = \lambda_\infty - b\sqrt{C}$

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Conductor and Conductance and Cell constant

At $298 \text{ K}$, the specific conductance $(\kappa)$ of a $0.0020 \text{ M } NaCl$ solution is $2.50 \times 10^{-4} \text{ S cm}^{-1}$. Calculate the molar conductivity $(\Lambda_m)$ of the solution in $\text{S cm}^2 \text{mol}^{-1}$.

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The molar conductivities of $KCl$,$NaCl$ and $KNO_3$ are $100$,$120$ and $90 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of $NaNO_3$ would be .......... $S \ cm^2 \ mol^{-1}$.

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Two solutions of electrolytes $X$ and $Y$ are taken in two beakers and diluted by adding $500 \ mL$ of water. The molar conductivity $(\Lambda_m)$ of $X$ increases by $1.5$ times,while that of $Y$ increases by $20$ times. What could be the electrolytes $X$ and $Y$?

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If the conductivity of mercury at $0^{\circ} \, C$ is $1.07 \times 10^{6} \, S \, m^{-1}$ and the resistance of a cell containing mercury is $0.243 \, \Omega$,then the cell constant of the cell is $x \times 10^{4} \, m^{-1}$. The value of $x$ is ...... (Nearest integer).

DifficultJEE MAIN 2021

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The resistance of a $1 \, N$ solution of a salt is $50 \, \Omega$. The two platinum electrodes in the solution are $2.1 \, cm$ apart and each has an area of $4.2 \, cm^2$. Calculate the equivalent conductivity of the solution.

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Which of the following is not a correct relationship between $\lambda_c$ and $\lambda_\infty$ for $NaCl$? Here,$\lambda_c$ and $\lambda_\infty$ are the equivalent conductance at concentration $C$ and at infinite dilution,respectively.

Similar Questions

At $298 \text{ K}$, the specific conductance $(\kappa)$ of a $0.0020 \text{ M } NaCl$ solution is $2.50 \times 10^{-4} \text{ S cm}^{-1}$. Calculate the molar conductivity $(\Lambda_m)$ of the solution in $\text{S cm}^2 \text{mol}^{-1}$.

The molar conductivities of $KCl$,$NaCl$ and $KNO_3$ are $100$,$120$ and $90 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of $NaNO_3$ would be .......... $S \ cm^2 \ mol^{-1}$.

Two solutions of electrolytes $X$ and $Y$ are taken in two beakers and diluted by adding $500 \ mL$ of water. The molar conductivity $(\Lambda_m)$ of $X$ increases by $1.5$ times,while that of $Y$ increases by $20$ times. What could be the electrolytes $X$ and $Y$?

If the conductivity of mercury at $0^{\circ} \, C$ is $1.07 \times 10^{6} \, S \, m^{-1}$ and the resistance of a cell containing mercury is $0.243 \, \Omega$,then the cell constant of the cell is $x \times 10^{4} \, m^{-1}$. The value of $x$ is ...... (Nearest integer).

The resistance of a $1 \, N$ solution of a salt is $50 \, \Omega$. The two platinum electrodes in the solution are $2.1 \, cm$ apart and each has an area of $4.2 \, cm^2$. Calculate the equivalent conductivity of the solution.

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