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
View Solution FreeMore Electrochemistry Questions
Cost of electricity for the production of ′X′ litre H2 at NTP at cathode is Rs. X, then cost of electricity ...For the given cell, pt, Cl2/HCl||HCl/Cl2, pt (l atm) C1 C2(1 atm) The emf will be positive when :-...At 298K the standard free energy of formation of H2O(l) is -237.20 kJ/mole while ᐃGo of its ionisation into H+ ion...Molar conductance of 0.1 M acetic acid is 7 ohm-1 cm2 mol-1. If the molar cond. of acetic acid at infinite dilution is 3...The standard potentials of MnO4−|Mn2+ and MnO2|Mn2+ electrodes in acid solution are 1.51 and 1.23 V, respectively. Stand...