of the solution using the provided molar conductivity Λ and concentration c.

Molar conductivity Λ is the conductivity per mole of solution and is related to the overall conductivity κ by the equation:

$$ \kappa = \Lambda \cdot c $$

Where:

κ is the conductivity in $$ \mathrm{ohm}^{-1} \mathrm{cm}^{-1} $$

Λ is the molar conductivity in $$ \mathrm{ohm}^{-1} \mathrm{cm}^2 \mathrm{mol}^{-1} $$

c is the concentration in moles per $$ \mathrm{cm}^3 $$

Given:

$$ \Lambda = 194.5 \, \mathrm{ohm}^{-1} \mathrm{cm}^2 \mathrm{mol}^{-1} $$

$$ c = 0.05 \, \mathrm{M} = 0.05 \, \mathrm{mol/L} = 0.05 \, \mathrm{mol/1000 \, cm}^3 = 0.00005 \, \mathrm{mol/cm}^3 $$

Calculating κ:

$$ \kappa = 194.5 \, \mathrm{ohm}^{-1} \mathrm{cm}^2 \mathrm{mol}^{-1} \times 0.00005 \, \mathrm{mol/cm}^3 = 0.009725 \, \mathrm{ohm}^{-1} \mathrm{cm}^{-1} $$

Next, we use the cell constant (G) to relate the resistance (R) of the cell to its conductivity. The cell constant is defined as the inverse of the distance between the electrodes divided by their area:

$$ G = \frac{l}{A} $$

Where:

l is the distance between the electrodes in cm

A is the area of the electrodes in $$ \mathrm{cm}^2 $$

Given:

$$ l = 1.0 \mathrm{~cm} $$

$$ A = 3.0 \mathrm{~cm}^2 $$

Calculating G:

$$ G = \frac{1.0 \, \mathrm{cm}}{3.0 \, \mathrm{cm}^2} = 0.3333 \, \mathrm{cm}^{-1} $$

The resistance R can then be calculated using the equation:

$$ R = \frac{1}{\kappa \cdot G} $$

$$ R = \frac{1}{0.009725 \, \mathrm{ohm}^{-1} \mathrm{cm}^{-1} \times 0.3333 \, \mathrm{cm}^{-1}} = 34.27 \, \mathrm{ohms} $$

Therefore, the resistance offered by the conductivity cell is 34.27 ohms.

The correct option is D (34.27 ohms).

Electrochemistry

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