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 of a 0.15 M solution of KCl at 298 K, given its conductivity (κ) is 0.0152 S/cm, we can use the formula for molar conductivity (Λm):

\[ Λ_m = \frac{κ \times 1000}{C} \]

where:
- \(Λ_m\) = molar conductivity (S cm²/mol)
- \(κ\) = conductivity (S/cm)
- \(C\) = concentration (mol/L or M)

### Step-by-Step Solution:

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

2. **Substitute the values into the formula:**
\[ Λ_m = \frac{0.0152 \, \text{S/cm} \times 1000}{0.15 \, \text{mol/L}} \]

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

4. **Now divide by the concentration:**
\[ Λ_m = \frac{15.2 \, \text{S cm}}{0.15 \, \text{mol/L}} = \frac{15.2}{0.15} \]

5. **Perform the division:**
\[ Λ_m = 101.33 \, \text{S cm}^2/\text{mol} \]

6. **Round to an appropriate number of significant figures:**
\[ Λ_m \approx 101 \, \text{S cm}^2/\text{mol} \]

### Final Answer:
The molar conductivity of the 0.15 M solution of KCl at 298 K is approximately **101 S cm²/mol**.