ElectrochemistryHard
Question
A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a conducivity cell at roomtemperature. What shall be the apporiximate molar condcutance of this NaOH solution if cell constant of the cell is 0.367 cm-1
Options
A.234 S cm2mole-1
B.23.2 S cm2mole-1
C.4645 S cm2mole-1
D.5464 S cm2mole-1
Solution
Here, R = 31.6ohm
∴ C =
ohm-1 = 0.0316 ohm-1
Specific conductance
= conductance × cell constant
= 0.0316ohm-1 × 0.367cm-1
= 0.0116ohm-1cm-1
Now ,molar concentration = 0.5 m (given)
= 0.5 × 10-3 mole cm-3
∴ Molar conductance =
=
= 23.2S cm2mol-1
∴ C =
ohm-1 = 0.0316 ohm-1Specific conductance
= conductance × cell constant
= 0.0316ohm-1 × 0.367cm-1
= 0.0116ohm-1cm-1
Now ,molar concentration = 0.5 m (given)
= 0.5 × 10-3 mole cm-3
∴ Molar conductance =
=
= 23.2S cm2mol-1Create a free account to view solution
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