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A. 0.05 M NaOH solution offered, a.resis...

A. `0.05 M NaOH` solution offered, a.resistance of 31.6 ohm in a conductivity cell at 298 K. if the cell constant is 0.367 cm^-1.calculate molar conductivity of the sodium hydroxide soIution.

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Cell-constant =0.367 cdot cm^x - Resistance, R `=31.6 .0 hm` Specific conductivity. `` = Cell constant / Resistanice =0.367/31.6=0.0116 ohm^-1 ~cm^prime . `` Concentration. `M=0.05 M` `therefore` Molar conductivity. `Lambda_m` ...
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