The equivalent conductances of two strong electrolytes at infinite dilution in H 2 O (where ions move freely through a solution) at 2 5 ∘ C are given below: ∧ ∘ CH 3 COONa = 91.0 S cm 2 /equiv ∧ H

Question

The equivalent conductances of two strong electrolytes at infinite dilution in $H_{2} O$ (where ions move freely through a solution) at $2 5^{\circ} C$ are given below: $\land^{\circ} CH_{3} COONa = 91.0 S cm^{2}$ /equiv $\land_{HCl}^{\circ} = 426.2 S c m^{2}$ / equiv What additional information/quantity one needs to calculate $\land^{\circ}$ of an aqueous solution of acetic acid?

Options:

Answer:

A

Solution:

From Kohlrausch's law $Λ_{CH_{3} COOH}^{\circ} = Λ_{CH_{3} COONa}^{\circ} + Λ_{HCl}^{\circ} - Λ_{NaCl}^{\circ}$ Hence, (A) is the correct answer.

TestHub · Accepted Answer

From Kohlrausch's law Λ CH 3 ​ COOH ∘ ​ = Λ CH 3 ​ COONa ∘ ​ + Λ HCl ∘ ​ − Λ NaCl ∘ ​ Hence, (A) is the correct answer.

Chemistry - Electrochemistry Question with Solution | TestHub

Options:

Doubts & Discussion