The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at `25^(@)C`are given below:
`wedge_(CH_(3)COONa^(@) = 91.0Scm^(2)//equ iv` , `wedge_HCl^(@)= 426.25cm^(2)//equiv`
What additional information/quantity one needs to calculate `overset(o)(wedge)` of an aqueous solution of acetic acid?

Options

A. `overset(o)(wedge)` of NaC1
B. `3 overset(o)(wedge)` of `CH_(3)COOK`
C. The limitation equivalent conductance of `H^(+) (wedge_(H^+)^(@))`
D. `overset(o)wedge` of chlorocetic acid `(C1CH_(2)COOH)`

Explanation

To calculate the equivalent conductance of an aqueous solution of acetic acid (CH₃COOH) at infinite dilution, we need to use the concept of equivalent conductance and the contributions from the ions involved. Here’s a step-by-step solution:

### Step 1: Understand the concept of equivalent conductance
Equivalent conductance (Λ) at infinite dilution is the sum of the conductances of the individual ions present in the electrolyte. For a weak electrolyte like acetic acid, we need to consider the ions it dissociates into.

### Step 2: Write the dissociation equation for acetic acid
Acetic acid (CH₃COOH) dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
...