At infinite dilution, the eq. conductances of `CH_(3)COONa`, HCl and `CH_(3)COOH` are 91, 426 and 391 mho `cm^(2)` respectively at `25^(@)C`. The eq. conductance of NaCl at infinite dilution will be

Options

A. 126
B. 209
C. 391
D. 908

Answer

To find the equivalent conductance of NaCl at infinite dilution, we can use the principle that at infinite dilution, the equivalent conductance of an electrolyte is the sum of the conductances of its individual ions.

### Step-by-Step Solution:

1. **Identify the given values**:
- Equivalent conductance of CH₃COONa (sodium acetate) = 91 mho cm²
- Equivalent conductance of HCl = 426 mho cm²
- Equivalent conductance of CH₃COOH (acetic acid) = 391 mho cm²

2. **Write the dissociation equations**:
- For CH₃COONa:
\[ \text{CH}_3\text{COONa} \rightarrow \text{CH}_3\text{COO}^- + \text{Na}^+ \]
Thus, \(\Lambda_{\text{CH}_3\text{COO}^-} + \Lambda_{\text{Na}^+} = 91\).

- For HCl:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Thus, \(\Lambda_{\text{H}^+} + \Lambda_{\text{Cl}^-} = 426\).

- For CH₃COOH (weak acid):
\[ \text{CH}_3\text{COOH} \rightarrow \text{H}^+ + \text{CH}_3\text{COO}^- \]
Thus, \(\Lambda_{\text{H}^+} + \Lambda_{\text{CH}_3\text{COO}^-} = 391\).

3. **Set up the equations**:
- From the dissociation of CH₃COONa:
\[ \Lambda_{\text{CH}_3\text{COO}^-} + \Lambda_{\text{Na}^+} = 91 \quad \text{(1)} \]
- From the dissociation of HCl:
\[ \Lambda_{\text{H}^+} + \Lambda_{\text{Cl}^-} = 426 \quad \text{(2)} \]
- From the dissociation of CH₃COOH:
\[ \Lambda_{\text{H}^+} + \Lambda_{\text{CH}_3\text{COO}^-} = 391 \quad \text{(3)} \]

4. **Solve the equations**:
- From equation (3), we can express \(\Lambda_{\text{H}^+}\):
\[ \Lambda_{\text{H}^+} = 391 - \Lambda_{\text{CH}_3\text{COO}^-} \quad \text{(4)} \]
- Substitute equation (4) into equation (2):
\[ (391 - \Lambda_{\text{CH}_3\text{COO}^-}) + \Lambda_{\text{Cl}^-} = 426 \]
Rearranging gives:
\[ \Lambda_{\text{Cl}^-} = 426 - 391 + \Lambda_{\text{CH}_3\text{COO}^-} = 35 + \Lambda_{\text{CH}_3\text{COO}^-} \quad \text{(5)} \]
- Substitute equation (5) into equation (1):
\[ \Lambda_{\text{CH}_3\text{COO}^-} + \Lambda_{\text{Na}^+} = 91 \]
Now we can express \(\Lambda_{\text{Na}^+}\):
\[ \Lambda_{\text{Na}^+} = 91 - \Lambda_{\text{CH}_3\text{COO}^-} \quad \text{(6)} \]

5. **Substitute and solve for \(\Lambda_{\text{Na}^+}\) and \(\Lambda_{\text{Cl}^-}\)**:
- Using equations (4) and (5) in (6):
\[ \Lambda_{\text{Na}^+} + (35 + \Lambda_{\text{CH}_3\text{COO}^-}) = 91 \]
- This leads to:
\[ (91 - \Lambda_{\text{CH}_3\text{COO}^-}) + (35 + \Lambda_{\text{CH}_3\text{COO}^-}) = 91 \]
- Simplifying gives:
\[ 126 = 91 \quad \text{(which is consistent)} \]

6. **Calculate the equivalent conductance of NaCl**:
- The equivalent conductance of NaCl at infinite dilution is:
\[ \Lambda_{\text{Na}^+} + \Lambda_{\text{Cl}^-} = (91 - \Lambda_{\text{CH}_3\text{COO}^-}) + (35 + \Lambda_{\text{CH}_3\text{COO}^-}) = 91 + 35 - \Lambda_{\text{CH}_3\text{COO}^-} + \Lambda_{\text{CH}_3\text{COO}^-} \]
- Thus:
\[ \Lambda_{\text{NaCl}} = 91 + 35 = 126 \text{ mho cm}^2 \]

### Final Answer:
The equivalent conductance of NaCl at infinite dilution is **126 mho cm²**.