On the basis of the given equivalent conductivity
`lamda_(oo)(NH_(4)Cl)=130`
`lamda_(oo)(OH^(-))=174`
`lamda_(oo)(Cl^(-))=66`
The value of `lamda_(oo)(NH_(4)OH)` will be

Options

A. 304
B. 238
C. 108
D. 64

Explanation

To find the equivalent conductivity of \( \lambda_{\infty}(NH_4OH) \), we can use the concept of equivalent conductivities of the ions involved in the dissociation of ammonium hydroxide.

1. **Identify the ions in \( NH_4OH \)**:
The dissociation of ammonium hydroxide can be represented as:
\[ NH_4OH \rightarrow NH_4^+ + OH^- \]
Therefore, the ions involved are \( NH_4^+ \) and \( OH^- \).

2. **Use the formula for equivalent conductivity**:
The equivalent conductivity of a compound can be calculated using the equivalent conductivities of its constituent ions:
\[ \lambda_{\infty}(NH_4OH) = \lambda_{\infty}(NH_4^+) + \lambda_{\infty}(OH^-) \]

3. **Substitute the known values**:
From the given data:
- \( \lambda_{\infty}(NH_4Cl) = 130 \)
- \( \lambda_{\infty}(OH^-) = 174 \)
- \( \lambda_{\infty}(Cl^-) = 66 \)

We need to find \( \lambda_{\infty}(NH_4^+) \). Since \( NH_4Cl \) dissociates into \( NH_4^+ \) and \( Cl^- \), we can express:
\[ \lambda_{\infty}(NH_4^+) = \lambda_{\infty}(NH_4Cl) - \lambda_{\infty}(Cl^-) \]
Substituting the values:
\[ \lambda_{\infty}(NH_4^+) = 130 - 66 = 64 \]

4. **Calculate \( \lambda_{\infty}(NH_4OH) \)**:
Now substitute \( \lambda_{\infty}(NH_4^+) \) and \( \lambda_{\infty}(OH^-) \) into the equation:
\[ \lambda_{\infty}(NH_4OH) = \lambda_{\infty}(NH_4^+) + \lambda_{\infty}(OH^-) \]
\[ \lambda_{\infty}(NH_4OH) = 64 + 174 = 238 \]

5. **Final Result**:
Therefore, the value of \( \lambda_{\infty}(NH_4OH) \) is:
\[ \lambda_{\infty}(NH_4OH) = 238 \]