A 0.1 M solution of monobasic acid at specific resistance of r ohms-cm, its molar conductivity is

Options

A. 10/r
B. 10r
C. `10^(4)//r`
D. `10^(4)r`

Answer

To find the molar conductivity of a 0.1 M solution of a monobasic acid with a specific resistance of R ohm-cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Molar conductivity (\( \Lambda_m \)) is defined as the conductivity (\( \kappa \)) multiplied by 1000 and divided by the molarity (\( n \)) of the solution.
- The formula for molar conductivity is given by:
\[ \Lambda_m = \frac{\kappa \times 1000}{n} \]

2. **Identify the Given Values**:
- Molarity (\( n \)) = 0.1 M
- Specific resistance (\( R \)) = R ohm-cm

3. **Calculate Conductivity**:
- Conductivity (\( \kappa \)) is the reciprocal of specific resistance (\( R \)):
\[ \kappa = \frac{1}{R} \]

4. **Substitute Conductivity into the Molar Conductivity Formula**:
- Now substitute the value of \( \kappa \) into the molar conductivity formula:
\[ \Lambda_m = \frac{\left(\frac{1}{R}\right) \times 1000}{0.1} \]

5. **Simplify the Expression**:
- Simplifying the expression gives:
\[ \Lambda_m = \frac{1000}{0.1 \times R} = \frac{10000}{R} \]

6. **Final Result**:
- Therefore, the molar conductivity of the solution is:
\[ \Lambda_m = \frac{10^4}{R} \text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1} \]

### Conclusion:
The molar conductivity of the 0.1 M solution of monobasic acid at specific resistance \( R \) ohm-cm is \( \frac{10^4}{R} \).