The equivalent conductance at infinite diluti | vedclass

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

Question

(A) According to Kohlrausch's law of independent migration of ions:$\wedge_{BA}^0 = \lambda_B^+ + \lambda_A^- = 140 \ S \ cm^2 \ eq^{-1}$$\wedge_{CA}^

Vedclass · Accepted Answer

$178$

Vedclass · Answer

(A) According to Kohlrausch's law of independent migration of ions:$\wedge_{BA}^0 = \lambda_B^+ + \lambda_A^- = 140 \ S \ cm^2 \ eq^{-1}$$\wedge_{CA}^0 = \lambda_C^+ + \lambda_A^- = 120 \ S \ cm^2 \ eq^{-1}$$\wedge_{BX}^0 = \lambda_B^+ + \lambda_X^- = 198 \ S \ cm^2 \ eq^{-1}$We need to find $\wedge_{CX}^0 = \lambda_C^+ + \lambda_X^-$.Using the relation: $\wedge_{CX}^0 = \wedge_{CA}^0 + \wedge_{BX}^0 - \wedge_{BA}^0$$\wedge_{CX}^0 = 120 + 198 - 140 = 178 \ S \ cm^2 \ eq^{-1}$

The equivalent conductance at infinite dilution $\wedge^0$ for electrolytes $BA$ and $CA$ are $140$ and $120 \ S \ cm^2 \ eq^{-1}$,respectively. The equivalent conductance at infinite dilution for $BX$ is $198 \ S \ cm^2 \ eq^{-1}$. The $\wedge^0$ (in $S \ cm^2 \ eq^{-1}$) of $CX$ is:

Similar Questions

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

If the resistivity of a $0.8 \ M \ KCl$ solution is $2.5 \times 10^{-3} \ \Omega \ cm$,calculate the molar conductivity of the solution.

What is electrical resistance? Give a note on it.

Calculate the molar conductivity of a solution having a concentration of $0.08 \ M$ and a resistivity of $5 \times 10^{-3} \ \Omega \ cm$.

What is the molar conductivity of $0.1 \ M$ $NaCl$ if its conductivity is $1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module