Dear Student, Resistance of cell, R= 245 ΩDistance between electrodes, l= 4cmArea of electrode, A= 7 cm2Now, Conductance, G=1R=1245Ω-1=0 004081 Ω-1and Conductivity, κ=G(lA)=0 004 ×47= 0 002332 Ω-1cm-1Now Molar Conductance, λm=κ×1000MFor KCl, Normality= Molarity ( as n-factor =1 )SO, Molarity of KCl solution=110MTherfore, λm=0 002332×1000110=23 32 Ω-1cm-1mol-1Hence, the correct answer is (3)

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Dear Student,

Resistance of cell, R = 245 Ω Distance between electrodes, l = 4 cm Area of electrode, A = 7 {cm}^{2} Now, Conductance, G = \frac{1}{R} = \frac{1}{245} Ω^{- 1} = 0.004081 Ω^{- 1} and Conductivity, κ = G (\frac{l}{A}) = 0.004 \times \frac{4}{7} = 0.002332 Ω^{- 1} {cm}^{- 1} Now Molar Conductance, λ_{m} = \frac{κ \times 1000}{M} For KCl, Normality = Molarity (as n - factor = 1) SO, Molarity of KCl solution = \frac{1}{10} M Therfore, λ_{m} = \frac{0.002332 \times 1000}{\frac{1}{10}} = 23.32 Ω^{- 1} {cm}^{- 1} {mol}^{- 1} Hence, the correct answer is (3)

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