For a strong electrolyte the molar conductivity follows Lambda_m = Lambda deg_m - Asqrtc with Lambda deg_m = 150.0 S cm^2 mol^-1 and A = 90.0 S cm^2 mol^-1 (mol L^-1)^-1/2. The conductivity of a 0.0100 M solution of this electrolyte is

JEE Main levelSingle correct optionDifficulty 2 of 3ec186

For a strong electrolyte the molar conductivity follows $Λ_{m} = Λ_{m}^{\circ} - A c$ with $Λ_{m}^{\circ} = 150.0 S c m^{2} mo l^{- 1}$ and $A = 90.0 S c m^{2} mo l^{- 1} (mol L^{- 1})^{- 1/2}$ . The conductivity of a $0.0100 M$ solution of this electrolyte is

c = 0.0100 = 0.100

, so

Λ_{m} = 150.0 - 90.0 \times 0.100 = 141.0 S c m^{2} mo l^{- 1}

. Then

κ = Λ_{m} c /1000 = 141.0 \times 0.0100/1000 = 1.41 \times 1 0^{- 3} S c m^{- 1}

. Using

Λ_{m}^{\circ}

directly gives

1.50 \times 1 0^{- 3}

; using

A c

instead of

A c

gives

1.49 \times 1 0^{- 3}