What is the molar conductance (in `S cm^(2) mol^(-1))` of 1M solution of acetic acid, if the resistance of that solution is 250 ohm and cell constant is equal to `1.15 cm^(-1)` ?

Options

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

Answer

To find the molar conductance of a 1M solution of acetic acid given the resistance and cell constant, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance (R) = 250 ohms
- Cell constant (k) = 1.15 cm⁻¹
- Concentration (C) = 1 M

2. **Calculate the Conductance (G):**
Conductance (G) is the reciprocal of resistance:
\[ G = \frac{1}{R} = \frac{1}{250 \, \text{ohms}} = 0.004 \, \text{S} \]

3. **Calculate the Conductivity (κ):**
The conductivity (κ) can be calculated using the formula:
\[ \kappa = G \times \text{cell constant} \]
Substituting the values:
\[ \kappa = 0.004 \, \text{S} \times 1.15 \, \text{cm}^{-1} = 0.0046 \, \text{S cm}^{-1} \]

4. **Calculate Molar Conductance (Λm):**
Molar conductance (Λm) is calculated using the formula:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
Substituting the values:
\[ \Lambda_m = \frac{0.0046 \, \text{S cm}^{-1} \times 1000}{1 \, \text{mol/L}} = 4.6 \, \text{S cm}^2 \text{mol}^{-1} \]

5. **Final Result:**
The molar conductance of the 1M solution of acetic acid is:
\[ \Lambda_m = 4.6 \, \text{S cm}^2 \text{mol}^{-1} \]