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0.05 M NaOH solution offered a resistance of 31.6 Ω in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm^-1, calculate the molar conductivity of NaOH solution. - Chemistry

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प्रश्न

0.05 M NaOH solution offered a resistance of 31.6 Ω in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm-1, calculate the molar conductivity of NaOH solution.

योग

उत्तर

Given:

Cell constant (b) = 0.367 cm–1
Molar concentration (C) = 0.05 M NaOH
Resistance (R) = 31.6 Ω

To find: Molar conductivity (^)

Solution:

m(NaOH) = ?

`k_((NaOH)) = b/R_((NaOH)) = 0.01161  Ω^-1 cm^-1`

`∧_(m(NaOH)) = (k xx 1000)/C`

`= (0.01161 xx 1000)/0.05`

= 232.2  Ω-1 cm2 mol-1

Molar conductivity = ∧m = 232.2  Ω-1 cm2 mol-1

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Electrical Conductance of Solution
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2016-2017 (March)

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