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Question

# The limiting molar conductivities Λ0 for NaCl, KBr and KCl are 126, 152 and 150 Scm2mol−1 respectively. The Λ0 for NaBr is:

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

## The correct option is A 128 Scm2mol−1 For a strong electrolyte NaBr, the limiting molar conductivity is the sum of the individual ionic conductivities. Λ0NaBr=λ0Na++λ0Br− Λ0NaBr=Λ0NaCl+Λ0KBr−Λ0KCl....eqn(1) Λ0NaBr=λ0Na++λ0Cl−+λ0K++λ0Br−−λ0Na+−λ0Br− Λ0NaBr=λ0Na++λ0Br− Hence, Substituting the values in equation (1), we get- Λ0NaBr=Λ0NaC1+Λ0KBr−Λ0KC1 Λ0NaBr=126+152−150 Λ0NaBr=128 Scm2mol−1 Hence, option A is correct.

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