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 NH₄OH at a given dilution. We will use the provided conductivities at infinite dilution and the equivalent conductance of the NH₄OH solution.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Conductivity at infinite dilution:
- λ(NH₄Cl) = 130 ohm⁻¹ cm² eq⁻¹
- λ(NaOH) = 218 ohm⁻¹ cm² eq⁻¹
- λ(NaCl) = 120 ohm⁻¹ cm² eq⁻¹
- Equivalent conductance of (N/100) solution of NH₄OH = 10 ohm⁻¹ cm² eq⁻¹

2. **Use Kohlrausch's Law:**
- According to Kohlrausch's Law, the equivalent conductance at infinite dilution of a weak electrolyte can be calculated using the conductance of its constituent ions.
- The dissociation of NH₄OH can be represented as:
\[ NH₄OH \rightleftharpoons NH₄^+ + OH^- \]
- Therefore, we can express the equivalent conductance of NH₄OH at infinite dilution (λ₀(NH₄OH)) as:
\[ λ₀(NH₄OH) = λ(NH₄^+) + λ(OH^-) - λ(Na^+) - λ(Cl^-) \]

3. **Calculate λ₀(NH₄OH):**
- Substitute the values into the equation:
\[ λ₀(NH₄OH) = λ(NH₄Cl) + λ(NaOH) - λ(NaCl) \]
- Plugging in the values:
\[ λ₀(NH₄OH) = 130 + 218 - 120 = 228 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

4. **Calculate Degree of Dissociation (α):**
- The degree of dissociation (α) can be calculated using the formula:
\[ α = \frac{λ_m}{λ₀} \]
- Where:
- λ_m = Equivalent conductance of the solution = 10 ohm⁻¹ cm² eq⁻¹
- λ₀ = Equivalent conductance at infinite dilution = 228 ohm⁻¹ cm² eq⁻¹
- Substitute the values:
\[ α = \frac{10}{228} \approx 0.04386 \]

5. **Final Result:**
- Therefore, the degree of dissociation of NH₄OH at this dilution is approximately **0.04386**.