ElectrochemistryEasy
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
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