The decreasing order of electrical conductivity of the following aqueous solution is :
0.1 M formic acid (A), 0.1 M Acetic acid (B), 0.1 M Benzoic acid (C).

Options

A. `CgtBgtA`
B. `CgtAgtB`
C. `AgtBgtC`
D. `AgtCgtB`

Explanation

To determine the decreasing order of electrical conductivity of the given aqueous solutions of formic acid (A), acetic acid (B), and benzoic acid (C), we need to consider the degree of ionization of each acid in solution. The conductivity of an aqueous solution is directly related to the concentration of ions present in the solution.

### Step-by-Step Solution:

1. **Identify the Acids and Their Properties**:
- Formic acid (HCOOH) - A
- Acetic acid (CH3COOH) - B
- Benzoic acid (C6H5COOH) - C

2. **Determine the Strength of Each Acid**:
- Formic acid is a stronger acid compared to acetic acid and benzoic acid.
- The pKa values are:
- Formic acid (pKa ≈ 3.75)
- Acetic acid (pKa ≈ 4.76)
- Benzoic acid (pKa ≈ 4.2)
- A lower pKa value indicates a stronger acid, which means it will ionize more in solution.

3. **Ionization of the Acids**:
- Stronger acids ionize more completely in solution, leading to a higher concentration of H⁺ ions.
- Formic acid (A) will produce more H⁺ ions than acetic acid (B) and benzoic acid (C).
- Benzoic acid (C) is stronger than acetic acid (B) but weaker than formic acid (A).

4. **Conductivity Ranking**:
- Since conductivity is related to the concentration of ions, the order of conductivity will be:
- Formic acid (A) > Benzoic acid (C) > Acetic acid (B)

5. **Final Answer**:
- The decreasing order of electrical conductivity is:
- **A > C > B**