For HCl solution at `25^(@)C,` equivalent conductance at infinite dilution, is `425 "ohm"​^(-1) cm​^(2) "equiv"^(-1)`​. The specific conductance of a solution of HCl is 0.3825 `"ohm​"^(-1) "​cm"​^(-1)`​. If the apparent degree of dissociation is 90% the normality of the solution is :-

Options

A. 0.90 N
B. 1.0 N
C. 10 N
D. 1.2 N

Explanation

To find the normality of the HCl solution given the specific conductance, equivalent conductance at infinite dilution, and the degree of dissociation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Equivalent conductance at infinite dilution (λ∞) = 425 ohm⁻¹ cm² equiv⁻¹
- Specific conductance (κ) = 0.3825 ohm⁻¹ cm⁻¹
- Degree of dissociation (α) = 90% = 0.9

2. **Calculate the Equivalent Conductance at the Given Concentration:**
The equivalent conductance (λ) at a particular concentration can be calculated using the formula:
\[ \lambda = \alpha \times \lambda_{\infty} \]
Substituting the values:
\[ \lambda = 0.9 \times 425 \, \text{ohm}^{-1} \text{cm}^2 \text{equiv}^{-1} = 382.5 \, \text{ohm}^{-1} \text{cm}^2 \text{equiv}^{-1} \]

3. **Relate Specific Conductance to Normality:**
The relationship between specific conductance (κ), equivalent conductance (λ), and normality (N) is given by:
\[ \lambda = \frac{1000 \kappa}{N} \]
Rearranging this formula to find normality (N):
\[ N = \frac{1000 \kappa}{\lambda} \]

4. **Substituting the Known Values:**
Now substitute κ and λ into the equation:
\[ N = \frac{1000 \times 0.3825}{382.5} \]

5. **Calculating Normality:**
\[ N = \frac{382.5}{382.5} = 1 \, \text{N} \]

### Final Answer:
The normality of the HCl solution is **1 N**.