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04/10/2020 10:13 am
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A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a conductivity cell at room temperature. What shall be the approximate molar conductance of this NaOH solution if cell constant of the cell is 0.367 cm-1.
(a) 234 S cm2 mole-1
(b) 23.2 S cm2 mole-1
(c) 4645 S cm2 mole-1
(d) 5464 S cm2 mole-1
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04/10/2020 10:23 am
(b) 23.2 S cm2 mole-1
Explanation:
Here, R = 31.6 ohm
∴ C = 1/R = 1/31.6 ohm-1 = 0.0316 ohm-1
Specific conductance
= conductance × cell constant
= 0.0316 ohm-1 x 0.367 cm-1
= 0.0116 ohm-1cm-1
Now, molar concentration = 0.5 M (given)
= 0.5 x 10-3 mole cm-3
∴ Molar conductance = k/molar conc.
= 0.0116/0.5 x 10-3
= 23.2 S cm2 mole-1