The specific conductance of a 0.01 M solution of KCl is 0.0014 `ohm^(-1) cm^(-1) " at " 25^(@)C`. Its equivalent conductance is:

Options

A. 14
B. 140
C. `1.4`
D. `0.14`

Answer

To find the equivalent conductance of a 0.01 M solution of KCl given its specific conductance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Specific conductance (κ) = 0.0014 ohm^(-1) cm^(-1)
- Molarity (C) = 0.01 M

2. **Understand the Relationship:**
- Molar conductance (Λ) is calculated using the formula:
\[ \Lambda = \frac{\kappa \times 1000}{C} \]
- Since KCl is a univalent electrolyte (valency = 1), the molarity is equal to normality. Therefore, equivalent conductance (Λeq) is the same as molar conductance (Λ).

3. **Substitute the Values into the Formula:**
- Substitute the specific conductance and molarity into the formula:
\[ \Lambda = \frac{0.0014 \, \text{ohm}^{-1} \text{cm}^{-1} \times 1000}{0.01} \]

4. **Calculate the Equivalent Conductance:**
- Calculate the numerator:
\[ 0.0014 \times 1000 = 1.4 \, \text{ohm}^{-1} \text{cm}^{-1} \]
- Now, divide by the molarity:
\[ \Lambda = \frac{1.4}{0.01} = 140 \, \text{ohm}^{-1} \text{cm}^{-1} \]

5. **Final Answer:**
- The equivalent conductance of the 0.01 M solution of KCl is **140 ohm^(-1) cm^(-1)**.