`wedge_m^0` for NaCl, HCl and NaAc are 120,400.2 and `80.5 Scm^2mol^-1` respectively calculate `wedge_m^0` for Hac

Options

A. `362.7Scm^2mol^-1`
B. `360.7Scm^2mol^-1`
C. `370.1Scm^2mol^-1`
D. `380.5Scm^2mol^-1`

Answer

To calculate the molar conductivity (`λ_m^0`) for acetic acid (HAc) using the provided values for NaCl, HCl, and NaAc, we will use the concept of Kohlrausch's law of independent ionic conductance. According to this law, the molar conductivity of an electrolyte can be expressed as the sum of the contributions from its constituent ions.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- `λ_m^0` for NaCl = 120 Scm²mol⁻¹
- `λ_m^0` for HCl = 400.2 Scm²mol⁻¹
- `λ_m^0` for NaAc = 80.5 Scm²mol⁻¹

2. **Write the Equation Based on Kohlrausch's Law**:
According to Kohlrausch's law, we can express the molar conductivity of acetic acid (HAc) as:
\[ λ_m^0(\text{HAc}) = λ_m^0(\text{HCl}) + λ_m^0(\text{NaAc}) - λ_m^0(\text{NaCl}) \]

3. **Substitute the Values into the Equation**:
\[ λ_m^0(\text{HAc}) = 400.2 + 80.5 - 120 \]

4. **Calculate the Result**:
- First, add the values:
\[ 400.2 + 80.5 = 480.7 \]
- Then, subtract the value for NaCl:
\[ 480.7 - 120 = 360.7 \]

5. **Final Result**:
\[ λ_m^0(\text{HAc}) = 360.7 \, \text{Scm}^2\text{mol}^{-1} \]

### Conclusion:
The molar conductivity of acetic acid (HAc) is `360.7 Scm²mol⁻¹`.