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Question

The limiting molar conductivities λ for NaCl,KBr and KCl are 126, 152 and 150 S cm2 mol1respectively. The λ for NaBr is

A
278 S cm2mol-1
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B
176 S cm2mol-1
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C
128 S cm2mol-1
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D
302 S cm2mol-1
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Solution

The correct option is C 128 S cm2mol-1

(126 scm2)NaCl=Na++Cl .......(1)
(152 scm2)KBr=K++Br .......(2)
(150 scm2)KCl=K++Cl .......(3)
By equation (1)+(2)(3)
NaBr=Na++Br=126+152150=128 Scm2mol1


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