Specific conductance of 0.1 M NaCl solution is `1.01xx10^(-2) ohm^(-1) cm^(-1)` . Its molar conductance in `ohm^(-1) cm^(2) mol^(-1)` is

Options

A. `1.01xx10^2`
B. `1.01xx10^3`
C. `1.01xx10^4`
D. `1.01`

Explanation

To find the molar conductance (Λm) of a 0.1 M NaCl solution given its specific conductance (κ), we can use the following formula:

\[ Λ_m = \frac{κ \times 1000}{C} \]

Where:
- \(Λ_m\) is the molar conductance in ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\),
- \(κ\) is the specific conductance in ohm\(^{-1}\) cm\(^{-1}\),
- \(C\) is the molarity of the solution in mol/L.

### Step-by-step solution:

1. **Identify the values**:
- Specific conductance (κ) = \(1.01 \times 10^{-2}\) ohm\(^{-1}\) cm\(^{-1}\)
- Molarity (C) = 0.1 M

2. **Substitute the values into the formula**:
\[ Λ_m = \frac{1.01 \times 10^{-2} \, \text{ohm}^{-1} \, \text{cm}^{-1} \times 1000}{0.1} \]

3. **Calculate the numerator**:
\[ 1.01 \times 10^{-2} \times 1000 = 10.1 \]

4. **Now divide by the molarity**:
\[ Λ_m = \frac{10.1}{0.1} = 101 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

5. **Final answer**:
\[ Λ_m = 101 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]