Calculte molar conductance at infinite dilution for acetic acid, given
`A_(m)^(oo)HCI=425 ohm^(-1)cm^(-1),A_(m)^(oo)NaCI=188 ohm^(-1)cm^(-1),A_(m)^(oo)CH_(3)COOHNa=96 ohm^(-1)cm^(-2)mol^(-1)`

Explanation

To calculate the molar conductance at infinite dilution for acetic acid (CH₃COOH), we can use the given data for hydrochloric acid (HCl), sodium chloride (NaCl), and the sodium acetate (CH₃COONa). The formula we will use is based on the principle of additivity of molar conductances.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Molar conductance at infinite dilution for HCl, \( A_m^{\infty} \text{(HCl)} = 425 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
- Molar conductance at infinite dilution for NaCl, \( A_m^{\infty} \text{(NaCl)} = 188 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
- Molar conductance at infinite dilution for sodium acetate, \( A_m^{\infty} \text{(CH}_3\text{COONa)} = 96 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)

2. **Set Up the Equation**:
We know that the molar conductance at infinite dilution for acetic acid can be calculated using the following relationship:
\[ A_m^{\infty} \text{(CH}_3\text{COOH)} = A_m^{\infty} \text{(CH}_3\text{COONa)} + A_m^{\infty} \text{(HCl)} - A_m^{\infty} \text{(NaCl)} \]

3. **Substitute the Values**:
Now, substitute the known values into the equation:
\[ A_m^{\infty} \text{(CH}_3\text{COOH)} = 96 + 425 - 188 \]

4. **Perform the Calculation**:
Calculate the right-hand side:
\[ A_m^{\infty} \text{(CH}_3\text{COOH)} = 96 + 425 = 521 \]
\[ A_m^{\infty} \text{(CH}_3\text{COOH)} = 521 - 188 = 333 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]

5. **Final Result**:
The molar conductance at infinite dilution for acetic acid is:
\[ A_m^{\infty} \text{(CH}_3\text{COOH)} = 333 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]