The conductivities at infinite dilution of `NH_(4)Cl, NaOH and NaCl` are 130, 218, 120 `ohm^(-1) cm^(2) eq^(-1)`. If equivalent conductance of `(N)/(100)` solution of `NH_(4)OH` is 10, then degree of dissociation of `NH_(4)OH` at this dilution is

Options

A. 0.005
B. 0.043
C. 0.01
D. 0.02

Answer

To solve the problem, we need to find the degree of dissociation (α) of \( NH_4OH \) at a dilution of \( \frac{N}{100} \) using the given conductivities at infinite dilution.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Conductivity at infinite dilution:
- \( \lambda_{NH_4Cl} = 130 \, \Omega^{-1} \, cm^2 \, eq^{-1} \)
- \( \lambda_{NaOH} = 218 \, \Omega^{-1} \, cm^2 \, eq^{-1} \)
- \( \lambda_{NaCl} = 120 \, \Omega^{-1} \, cm^2 \, eq^{-1} \)
- Equivalent conductance of \( \frac{N}{100} \) solution of \( NH_4OH \) is given as \( \Lambda = 10 \, \Omega^{-1} \, cm^2 \, eq^{-1} \).

2. **Calculate the Limiting Equivalent Conductance of \( NH_4OH \):**
- The limiting equivalent conductance \( \lambda_{NH_4OH} \) can be calculated using the contributions from the ions \( NH_4^+ \) and \( OH^- \):
\[ \lambda_{NH_4OH} = \lambda_{NH_4^+} + \lambda_{OH^-} \]
- From the given data:
- \( \lambda_{NH_4^+} = \lambda_{NH_4Cl} = 130 \, \Omega^{-1} \, cm^2 \, eq^{-1} \)
- \( \lambda_{OH^-} = \lambda_{NaOH} = 218 \, \Omega^{-1} \, cm^2 \, eq^{-1} \)

- Therefore,
\[ \lambda_{NH_4OH} = 130 + 218 = 348 \, \Omega^{-1} \, cm^2 \, eq^{-1} \]

3. **Use the Formula for Degree of Dissociation:**
- The degree of dissociation \( \alpha \) is given by the formula:
\[ \alpha = \frac{\Lambda}{\lambda_{NH_4OH}} \]
- Where \( \Lambda \) is the equivalent conductance of the solution at a certain dilution.

4. **Substitute the Values:**
- Substitute \( \Lambda = 10 \, \Omega^{-1} \, cm^2 \, eq^{-1} \) and \( \lambda_{NH_4OH} = 348 \, \Omega^{-1} \, cm^2 \, eq^{-1} \):
\[ \alpha = \frac{10}{348} \]

5. **Calculate the Degree of Dissociation:**
- Performing the calculation:
\[ \alpha = 0.0287 \approx 0.029 \]

### Final Answer:
The degree of dissociation of \( NH_4OH \) at this dilution is approximately \( 0.029 \) or \( 2.9\% \).