The resistance of `0.1N` solution of formic acid is `200 ohm` and cell constant is `2.0 cm^(-1)`. The equivalent conductivity ( in `Scm^(2) eq^(-1))` of `0.1N` formic acid is `:`

Options

A. 100
B. 10
C. 1
D. none of these

Explanation

To find the equivalent conductivity of a 0.1N solution of formic acid, we can follow these steps:

### Step 1: Write down the given data
- Resistance (R) = 200 ohms
- Cell constant (G*) = 2.0 cm^(-1)
- Normality (N) = 0.1 N

### Step 2: Calculate the conductivity (K)
The conductivity (K) can be calculated using the formula:
\[ K = \frac{1}{R} \times G^* \]
Where:
- \( R \) is the resistance in ohms.
- \( G^* \) is the cell constant in cm^(-1).

Substituting the values:
\[ K = \frac{1}{200} \times 2.0 \]
\[ K = 0.01 \, \text{S cm}^{-1} \]

### Step 3: Calculate the equivalent conductivity (Λ)
The equivalent conductivity (Λ) is given by the formula:
\[ \Lambda = K \times \frac{1000}{N} \]
Where:
- \( K \) is the conductivity in S cm^(-1).
- \( N \) is the normality.

Substituting the values:
\[ \Lambda = 0.01 \times \frac{1000}{0.1} \]
\[ \Lambda = 0.01 \times 10000 \]
\[ \Lambda = 100 \, \text{S cm}^2 \text{eq}^{-1} \]

### Final Answer
The equivalent conductivity of 0.1N formic acid is:
\[ \Lambda = 100 \, \text{S cm}^2 \text{eq}^{-1} \]