resistance ($$R$$) of the solution using Ohm's Law:

$$R = \frac{V}{I}$$

where $$V$$ is the potential difference (40 V) and $$I$$ is the current (0.4 A).

2. Calculate the conductivity ($$\kappa$$) using the relationship:

$$\kappa = \frac{1}{R} \cdot \frac{l}{A}$$

where $$l$$ is the length of the column (12 cm) and $$A$$ is the cross-sectional area of the column. The area $$A$$ can be calculated using the diameter ($$d$$) as follows:

$$A = \pi \left(\frac{d}{2}\right)^2$$

3. Finally, calculate the molar conductivity ($$\Lambda_m$$) using the formula:

$$\Lambda_m = \frac{\kappa}{C}$$

where $$C$$ is the concentration of the solution (0.1 M).

Let's go through the steps one by one:

Step 1: Calculate the resistance ($$R$$)

$$R = \frac{V}{I} = \frac{40 \, \text{V}}{0.4 \, \text{A}} = 100 \, \Omega$$

Step 2: Calculate the cross-sectional area ($$A$$)

$$d = 4.0 \, \text{cm} \Rightarrow \text{radius} = \frac{d}{2} = 2.0 \, \text{cm}$$

$$A = \pi \left(2.0 \, \text{cm}\right)^2 = 4\pi \, \text{cm}^2$$

Step 3: Calculate the conductivity ($$\kappa$$)

$$\kappa = \frac{1}{R} \cdot \frac{l}{A} = \frac{1}{100 \, \Omega} \cdot \frac{12 \, \text{cm}}{4\pi \, \text{cm}^2} = \frac{12}{400\pi} \, \text{S/cm}$$

$$\kappa = \frac{12}{400\pi} \, \text{S/cm} = \frac{12}{1256.64} \, \text{S/cm} \approx 0.009546 \, \text{S/cm}$$

Step 4: Calculate the molar conductivity ($$\Lambda_m$$)

$$\Lambda_m = \frac{\kappa}{C} = \frac{0.009546 \, \text{S/cm}}{0.1 \, \text{M}} = 0.09546 \, \text{S cm}^2 \, \text{mol}^{-1}$$

Matching the calculated value with the options provided, the closest option is:

Option B: $$95.5 \, \text{Scm}^2 \, \text{mol}^{-1}$$

Electrochemistry

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