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

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 of a 0.1 M NaCl solution given its specific conductance, we can follow these steps:

### Step 1: Understand the relationship between specific conductance and molar conductance
Molar conductance (Λ) is defined as the specific conductance (κ) divided by the molar concentration (C) of the solution. The formula is:
\[ \Lambda = \frac{\kappa}{C} \]

### Step 2: Identify the given values
From the question, we have:
- Specific conductance (κ) = \(1.06 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1}\)
- Molar concentration (C) = 0.1 M

### Step 3: Substitute the values into the formula
Using the formula for molar conductance:
\[ \Lambda = \frac{1.06 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1}}{0.1 \, \text{mol/L}} \]

### Step 4: Calculate the molar conductance
Perform the division:
\[ \Lambda = \frac{1.06 \times 10^{-2}}{0.1} = 1.06 \times 10^{-1} \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Step 5: Convert the units if necessary
Since \(1.06 \times 10^{-1} \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1}\) can also be expressed as:
\[ \Lambda = 1.06 \times 10^{-1} \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Final Answer
Thus, the molar conductance of the 0.1 M NaCl solution is:
\[ \Lambda = 1.06 \times 10^{-1} \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

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