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Question 85 of 111

Q.0.05 M NaOH solution offered a resistance of 31.6 Ω31.6\ \Omega in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm−10.367\ cm^{-1}, calculate the molar conductivity of NaOH solution.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 3mImportance★★★★★
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Specific conductance from cell-constant/resistance, then scaled to molar conductance using the molarity.

Step 1 — specific conductance (κ\kappa):

κ=cell constantR=0.367 cm−131.6 Ω=0.01161 S cm−1\kappa = \dfrac{\text{cell constant}}{R} = \dfrac{0.367\ cm^{-1}}{31.6\ \Omega} = 0.01161\ S\,cm^{-1}

Step 2 — molar conductance (Λm\Lambda_m):

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