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Question

The limiting molar conductivities of NaCl, KBr and KCl are 126, 152 and 150 Scm2mol1 respectively. The limiting molar conductivity for NaBr is:

A
302 Scm2mol1
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B
176 Scm2mol1
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C
278 Scm2mol1
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D
128 Scm2mol1
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Solution

The correct option is D 128 Scm2mol1
The limiting molar conductivities of NaCl, KBr and KCl are 126, 152 and 150 Scm2mol1 respectively.

The limiting molar conductivity Λo for NaBr is calculated by the following expression.

λNaBr=λNaCl+λKBrλKCl

λNaBr=126+152150=128 Scm2mol1

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