Molar conductivity of 0.15 M solution of KCI at 298 K, if its conductivity is 0.0152 S `cm^(-1)` will be

Options

A. `124 Omega^(-1) "cm"^(2) "mol"^(-1)`
B. `204 Omega^(-1) "cm"^(2) "mol"^(-1)`
C. `101 Omega^(-1) "cm"^(2) "mol"^(-1)`
D. `300 Omega^(-1) "cm"^(2) "mol"^(-1)`

Explanation

To find the molar conductivity (Λm) of a 0.15 M solution of KCl at 298 K given its conductivity (κ) of 0.0152 S cm⁻¹, we can use the formula:

\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]

Where:
- \( \Lambda_m \) = molar conductivity in S cm² mol⁻¹
- \( \kappa \) = conductivity in S cm⁻¹
- \( C \) = concentration in mol L⁻¹ (M)

### Step-by-Step Solution:

1. **Identify the given values**:
- Conductivity (κ) = 0.0152 S cm⁻¹
- Concentration (C) = 0.15 M

2. **Convert the concentration to the appropriate units**:
- Since the concentration is already in mol L⁻¹, we can use it directly.

3. **Substitute the values into the formula**:
\[ \Lambda_m = \frac{0.0152 \, \text{S cm}^{-1} \times 1000}{0.15 \, \text{mol L}^{-1}} \]

4. **Calculate the numerator**:
\[ 0.0152 \times 1000 = 15.2 \, \text{S cm}^{-1} \]

5. **Now divide by the concentration**:
\[ \Lambda_m = \frac{15.2 \, \text{S cm}^{-1}}{0.15 \, \text{mol L}^{-1}} = 101.33 \, \text{S cm}^2 \text{mol}^{-1} \]

6. **Round off the answer**:
- The molar conductivity of the solution is approximately \( 101.33 \, \text{S cm}^2 \text{mol}^{-1} \).

### Final Answer:
\[ \Lambda_m \approx 101.33 \, \text{S cm}^2 \text{mol}^{-1} \]