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. 18.4
B. 9.2
C. 4.6
D. 2.3

Answer

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

### Step 1: Identify the given values
- Resistance (R) of the solution = 250 ohms
- Cell constant (L/A) = 1.15 cm⁻¹
- Normality (N) of the solution = 1N

### Step 2: Calculate the conductivity (κ)
The conductivity (κ) can be calculated using the formula:
\[ \kappa = \frac{L}{R} \]
Substituting the values:
\[ \kappa = \frac{1.15 \, \text{cm}^{-1}}{250 \, \text{ohm}} = 0.0046 \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Calculate the equivalent conductance (Λ)
The equivalent conductance (Λ) can be calculated using the formula:
\[ \Lambda = \frac{\kappa \times 1000}{N} \]
Substituting the values:
\[ \Lambda = \frac{0.0046 \, \text{ohm}^{-1} \text{cm}^{-1} \times 1000}{1} = 4.6 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

### Final Answer
The equivalent conductance of 1N acetic acid is \( 4.6 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \).

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