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Question

# The limiting molar conductivities of NaCl, KBr and KCl are 126, 152 and 150 Scm2molâˆ’1 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 Scm2mol−1The limiting molar conductivities of NaCl, KBr and KCl are 126, 152 and 150 Scm2mol−1 respectively. The limiting molar conductivity Λo for NaBr is calculated by the following expression.λ∞NaBr=λ∞NaCl+λ∞KBr−λ∞KClλ∞NaBr=126+152−150=128 Scm2mol−1

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