The molar conductivity of a strong electrolyte is 135.5 S cm^2 mol^-1 at c = 0.0025 M and 126.5 S cm^2 mol^-1 at c = 0.0225 M. Assuming the relation Lambda_m = Lambda deg_m - Asqrtc holds over the whole range, the conductivity of a 0.160 M solution of this electrolyte, in mS cm^-1 (two decimals), is

JEE Advanced levelNumerical answerDifficulty 3 of 3ec188

The molar conductivity of a strong electrolyte is $135.5 S c m^{2} mo l^{- 1}$ at $c = 0.0025 M$ and $126.5 S c m^{2} mo l^{- 1}$ at $c = 0.0225 M$ . Assuming the relation $Λ_{m} = Λ_{m}^{\circ} - A c$ holds over the whole range, the conductivity of a $0.160 M$ solution of this electrolyte, in $mS c m^{- 1}$ (two decimals), is

Numerical answer type, answer in mS cm^-1, tolerance 0.05.

0.0025 = 0.050

and

0.0225 = 0.150

. Subtracting the two equations:

A (0.150 - 0.050) = 135.5 - 126.5 = 9.0

, so

A = 90.0

and

Λ_{m}^{\circ} = 135.5 + 90.0 \times 0.050 = 140.0 S c m^{2} mo l^{- 1}

. At

c = 0.160

c = 0.400

and

Λ_{m} = 140.0 - 90.0 \times 0.400 = 104.0 S c m^{2} mo l^{- 1}

κ = Λ_{m} c /1000 = 104.0 \times 0.160/1000 = 1.664 \times 1 0^{- 2} S c m^{- 1} = 16.64 mS c m^{- 1}