ElectrochemistryHard
Question
The limiting molar conductivities ∧o for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol-1 respectively. The ∧o for NaBr is
Options
A.128 S cm2 mol-1
B.302 S cm2 mol-1
C.278 S cm2 mol-1
D.176 S cm2 mol-1
Solution
∧oNaCl = λoCl = 126 .....(1)
∧oKBr = λoK++ λoBr- = 152 .....(2)
∧oKCl = λoK++ λoCl- = 150 .....(3)
∧oNaBr = λoNa + λoBr-
∧oNaBr = 126 + 152 - 150 = 128
∧oKBr = λoK++ λoBr- = 152 .....(2)
∧oKCl = λoK++ λoCl- = 150 .....(3)
∧oNaBr = λoNa + λoBr-
∧oNaBr = 126 + 152 - 150 = 128
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
How many faradays are required to reduce one mol of MnO4- to Mn2+ -...The overall formation constant for the reaction of 6 mole of CN− with cobalt (II) is 1 × 1019. What is the formation con...The oxidation potentials of Zn, Cu, Ag, H2 and Ni are 0.76, -0.34, -0.80, 0.00, 0.25 volt, respectively. Which of the fo...What is the amount of chlorine evolved, when 2 amp of current is passed for 30 minutes in an aqueous solution of NaCl?...Any redox reaction would occur spontaneously, if :...