Which of the following expressions correctly | vedclass

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

Question

(A) The equivalent conductance at infinite dilution $(\Lambda_{eq}^{\infty})$ of an electrolyte is the sum of the equivalent conductances of its const

Vedclass · Accepted Answer

$\Lambda_{Al^{3+}}^o + \Lambda_{SO_4^{2-}}^o$

Vedclass · Answer

(A) The equivalent conductance at infinite dilution $(\Lambda_{eq}^{\infty})$ of an electrolyte is the sum of the equivalent conductances of its constituent ions.
By definition,the equivalent conductance of any electrolyte at infinite dilution is the sum of the equivalent conductances of its cation and anion.
For $Al_2(SO_4)_3$,the equivalent conductance at infinite dilution is given by:
$\Lambda_{eq}^{\infty} = \Lambda_{Al^{3+}}^o + \Lambda_{SO_4^{2-}}^o$
This is because the equivalent conductance of an ion is defined as the molar conductance divided by its charge (valence factor),which inherently accounts for the stoichiometry of the salt.

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of $Al_2(SO_4)_3$? Given that $\Lambda_{Al^{3+}}^o$ and $\Lambda_{SO_4^{2-}}^o$ are the equivalent conductances at infinite dilution of the respective ions.

Similar Questions

Discuss the construction and uses of a conductivity cell.

$A$ conductivity cell shows a resistance of $600 \ \Omega$. If the conductivity of $0.01 \ M \ KCl$ is $0.0015 \ \Omega^{-1} \ cm^{-1}$,what is the cell constant (in $cm^{-1}$)?

The limiting molar conductivities $(\Lambda_0)$ for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ S \ cm^2 \ mol^{-1}$ respectively. What is the $\Lambda_0$ of $NaBr$?

The conductivity of $0.02 \ M$ $KCl$ solution is $0.00250 \ \Omega^{-1} \ cm^{-1}$. What is its molar conductivity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module