The conductivity of 0.01 `mol L^(-1)`KCl solution is `1.41xx10^(-3)" S "cm^(-1)`. What is the molar conductivity (`S cm^(2) mol^(-1)`) ?

Answer

To find the molar conductivity of the KCl solution, we can use the formula for molar conductivity (\( \Lambda_m \)):

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

where:
- \( K \) is the conductivity of the solution in \( S \, cm^{-1} \)
- \( C \) is the molarity of the solution in \( mol \, L^{-1} \)

### Step-by-Step Solution:

1. **Identify the given values**:
- Conductivity \( K = 1.41 \times 10^{-3} \, S \, cm^{-1} \)
- Molarity \( C = 0.01 \, mol \, L^{-1} \)

2. **Substitute the values into the formula**:
\[ \Lambda_m = \frac{1000 \times (1.41 \times 10^{-3})}{0.01} \]

3. **Calculate the numerator**:
\[ 1000 \times 1.41 \times 10^{-3} = 1.41 \]

4. **Divide by the molarity**:
\[ \Lambda_m = \frac{1.41}{0.01} = 141 \, S \, cm^{2} \, mol^{-1} \]

5. **Final result**:
The molar conductivity of the KCl solution is \( 141 \, S \, cm^{2} \, mol^{-1} \).