The resistance of `1N` solution of acetic acid is `250ohm`, when measured in a cell of cell constant `1.15 cm^(-1)`. The equivalent conductance `(` in `ohm^(-1) cm^(2)eq^(-1))` of `1N` acetic acid is

Options

A. `4.6`
B. `9.2`
C. `18.4`
D. `2.3`

Explanation

To find the equivalent conductance of a 1N solution of acetic acid, we will follow these steps:

### Step 1: Understand the relationship between resistance, cell constant, and specific conductance.
The specific conductance (K) can be calculated using the formula:
\[ K = \frac{Cell \, Constant}{Resistance} \]

### Step 2: Substitute the given values into the formula.
We know from the problem:
- Resistance (R) = 250 ohms
- Cell constant (C) = 1.15 cm\(^{-1}\)

Substituting these values into the formula gives:
\[ K = \frac{1.15 \, \text{cm}^{-1}}{250 \, \Omega} \]

### Step 3: Calculate the specific conductance (K).
Now, we perform the calculation:
\[ K = \frac{1.15}{250} = 0.0046 \, \text{cm}^{-1} \]

### Step 4: Calculate the equivalent conductance (Λ).
The equivalent conductance (Λ) is given by the formula:
\[ \Lambda = K \times \frac{1000}{C} \]
where C is the concentration in gram equivalents per liter. Since we have a 1N solution, C = 1.

Substituting the values:
\[ \Lambda = 0.0046 \, \text{cm}^{-1} \times \frac{1000}{1} \]

### Step 5: Perform the calculation for equivalent conductance.
Calculating gives:
\[ \Lambda = 0.0046 \times 1000 = 4.6 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

### Final Answer:
The equivalent conductance of 1N acetic acid is:
\[ \Lambda = 4.6 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

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