For an aqueous solution of CaCl_2 electrolyte | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The variation of molar conductivity $(\lambda_m)$ with concentration $(C)$ for strong electrolytes is given by the Kohlrausch equation: $\lambda_m

Vedclass · Accepted Answer

Vedclass · Answer

(B) The variation of molar conductivity $(\lambda_m)$ with concentration $(C)$ for strong electrolytes is given by the Kohlrausch equation: $\lambda_m = \lambda_m^\circ - A\sqrt{C}$.Here,$\lambda_m^\circ$ is the molar conductivity at infinite dilution,$A$ is a constant,and $\sqrt{C}$ is the square root of the concentration.This equation represents a straight line equation of the form $y = mx + c$,where $y = \lambda_m$,$x = \sqrt{C}$,$m = -A$ (slope),and $c = \lambda_m^\circ$ (intercept).Since the slope is negative $(-A)$,the graph of $\lambda_m$ versus $\sqrt{C}$ is a straight line with a negative slope.

For an aqueous solution of $CaCl_2$ electrolyte,the graph between molar conductivity $(\lambda_m)$ and $(\text{concentration})^{1/2}$ is:

Similar Questions

If the specific conductivity of $0.1 \, N$ nitric acid is $6.2 \times 10^{-2} \, \Omega^{-1} \, cm^{-1}$,what will be the molar conductivity of the solution in $\Omega^{-1} \, cm^{2} \, mol^{-1}$?

Molar conductivity of $0.04 \ M \ BaCl_2$ solution is $230 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ at $27^{\circ} C$. What is its conductivity?

What is the common unit of conductivity if the dimensions are expressed in centimeter?

The unit $ohm^{-1}$ is used for

The molar conductivity of $0.02 \ M$ $KCl$ solution is $410 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ at $25^{\circ} C$. Calculate its conductivity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module