The specific conductance of 0.1 M NaCl solution is `1.06 xx 10^(-2) ohm^(-1) cm^(-1)` . Its molar conductance in `ohm^(-1) cm^(2) mol^(-1)` is

Options

A. `1.06 xx 10^2`
B. `1.06 xx 10^(3)`
C. `1.06 xx 10^(4)`
D. `53`

Explanation

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

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

where:
- \( \Lambda_m \) is the molar conductance in \( \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \)
- \( \kappa \) is the specific conductance in \( \text{ohm}^{-1} \text{cm}^{-1} \)
- \( C \) is the concentration in mol/L (M)

### Step-by-Step Solution:

1. **Identify the given values**:
- Specific conductance (\( \kappa \)) = \( 1.06 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1} \)
- Concentration (\( C \)) = \( 0.1 \, \text{mol/L} \)

2. **Substitute the values into the formula**:
\[ \Lambda_m = \frac{1.06 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1} \times 1000}{0.1} \]

3. **Calculate the numerator**:
\[ 1.06 \times 10^{-2} \times 1000 = 1.06 \times 10^{1} = 10.6 \, \text{ohm}^{-1} \text{cm}^{-1} \]

4. **Divide by the concentration**:
\[ \Lambda_m = \frac{10.6 \, \text{ohm}^{-1} \text{cm}^{-1}}{0.1} = 106 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

5. **Express the final answer in scientific notation**:
\[ \Lambda_m = 1.06 \times 10^{2} \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Final Answer:
The molar conductance of the 0.1 M NaCl solution is \( 1.06 \times 10^{2} \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \).

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