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Question

The limiting molar conductivities Λ0
for NaCl, KBr and KCl are 126, 152 and 150 Scm2mol1 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 Scm2mol1

For a strong electrolyte NaBr, the limiting molar conductivity is the sum of the individual ionic conductivities.

Λ0NaBr=λ0Na++λ0Br

Λ0NaBr=Λ0NaCl+Λ0KBrΛ0KCl....eqn(1)

Λ0NaBr=λ0Na++λ0Cl+λ0K++λ0Brλ0Na+λ0Br

Λ0NaBr=λ0Na++λ0Br

Hence,
Substituting the values in equation (1), we get-

Λ0NaBr=Λ0NaC1+Λ0KBrΛ0KC1

Λ0NaBr=126+152150
Λ0NaBr=128 Scm2mol1

Hence, option A is correct.


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