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 (Λm) 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 (Λm) is defined as the conductivity (κ) of the solution multiplied by 1000 and divided by the molarity (C) of the solution.
- The relationship is given by the formula:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]

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

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

4. **Substitute Conductivity in the Molar Conductivity Formula**:
- Substitute κ in the molar conductivity formula:
\[ \Lambda_m = \frac{\left(\frac{1}{R}\right) \times 1000}{0.1} \]

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

6. **Final Expression**:
- Thus, the molar conductivity (Λm) of the solution is:
\[ \Lambda_m = \frac{10000}{R} \text{ S cm}^2/\text{mol} \]

### Final Answer:
The molar conductivity of a 0.1 M solution of monobasic acid at specific resistance R ohm-cm is:
\[ \Lambda_m = \frac{10000}{R} \text{ S cm}^2/\text{mol} \]