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 of \(4\, \ cm^2\) will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(62.5\, \ Scm^2mol^{-1}\)
The conductivity of a solution is found from its resistance and the cell constant G* = l/A of the cell:
κ = G*/R = (l/A) × (1/R)
and the molar conductivity from the conductivity and the molarity M:
Λm = κ × 1000/M (κ in S cm−1, 1000 cm3 L−1)
Here l = 2 cm, A = 4 cm2, R = 40 Ω and M = 0.2 mol L−1:
G* = 2 cm / 4 cm2 = 0.5 cm−1
κ = 0.5 cm−1 / 40 Ω = 0.0125 S cm−1
Λm = 0.0125 × 1000 / 0.2 = 62.5 S cm2 mol−1
NCERT (2026-27 edition):
Measurement of the Conductivity of Ionic Solutions – “… If we use S cm–1 as the units for κ and mol cm–3, the units of concentration, then the units for Λm are S cm2 mol–1. It can be calculated by using the equation:”
\( \displaystyle \Lambda_m\ (\mathrm{S\,cm^2\,mol^{-1}}) = \frac{\kappa\ (\mathrm{S\,cm^{-1}}) \times 1000\ (\mathrm{cm^3/L})}{\text{molarity}\ (\mathrm{mol/L})} \)