Answer the question on basis of passage given below:

\(G = \frac{\kappa A}{l} = \kappa \) (both\(A\) and \(l\) are unity in their appropriate units in m or cm)

Molar conductivity of a solution at a given concentration is the conductance of the volume \(V\) of solution containing one mole of electrolyte kept between two electrodes with area of cross section \(A\) and distance of unit length \(l\). Therefore,

\(\Lambda_m = \frac{\kappa A}{l}\)

Since \(l = 1\) and \(A = V\) (volume containing 1 mole of electrolyte) \(\Lambda _m =\kappa V\)

Molar conductance for \(0.2\, \ M\, \ AgCl\) solution showing resistance of \(40\, \ \Omega\) in a cell having distance between electrodes as \(2\, \ cm\) and surface area area of \(4\, \ cm^2\) will be:

\(62.5\, \ Scm^2mol^{-1}\)

The correct answer is option 3. \(62.5\, \ Scm^2mol^{-1}\).

To calculate the molar conductance (\(\Lambda_m\)) for the given \(0.2\, M\) \(AgCl\) solution, we can follow these steps:

The conductance \(G\) is the reciprocal of resistance \(R\):

\(G = \frac{1}{R}\)

Given that the resistance \(R = 40\, \Omega\):

\(G = \frac{1}{40}\, S = 0.025\, S\)

Conductivity \(\kappa\) is given by:

\(\kappa = G \times \frac{l}{A}\)

Where:

\(G\) is the conductance,

\(l = 2\, \text{cm}\) is the distance between the electrodes,

\(A = 4\, \text{cm}^2\) is the area of the electrodes.

Substituting the values:

\(\kappa = 0.025\, S \times \frac{2\, \text{cm}}{4\, \text{cm}^2} = 0.0125\, S/cm\)

Molar conductance \(\Lambda_m\) is calculated using the formula:

\(\Lambda_m = \frac{\kappa \times 1000}{C}\)

Where:

\(\kappa\) is the conductivity,

\(C = 0.2\, M\) is the concentration.

Substituting the values:

\(\Lambda_m = \frac{0.0125 \times 1000}{0.2} = \frac{12.5}{0.2} = 62.5\, Scm^2/mol\)

The molar conductance \(\Lambda_m\) is \(62.5\, Scm^2/mol\).

So, the correct answer is: 3. \(62.5\, \ Scm^2mol^{-1}\)