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Dear Student,

Please find below the solution to the asked query:

1) Calculate the cell constant, K:

K = specific conductivity (κ) x Resistance (R) = (1.29 x 10^-2 ohm^-1 cm^-1) x (85 ohm) = 1.0965 cm^-1

2) Using this cell constant calculate the specific conductivity of the electrolyte at 0.052 M concentration.

specific conductivity (κ) = K / R = 1.0965 cm^-1 / 96 ohm = 1.14 x 10^-2 ohm^-1 cm^-1

3) Molar conductivity = $$\begin{gathered} \frac{Specific conductivity (oh{m}^{-1} c{m}^{-1})}{concentration (mol L^{-1})}\times 1000 (c{m}^3 L^{-1}) \\ = \frac{1.14\times {10}^{-2} oh{m}^{-1} c{m}^{-1}}{0.052 mol L^{-1}}\times 1000 (c{m}^3 L^{-1} \\ = \mathbf{219}\mathbf{ }\mathit{o}\mathit{h}{\mathit{m}}^{\mathbf{-}\mathbf{1}}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}\mathbf{ }\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Hence, the molar conductivity of the unknown electrolyte at 0.052 M is 219 ohm^-1 cm² mol^-1.

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