The limiting molar conductivities (Λ0) for NaCl, KBr and KCl are 126, 152 and150 Scm2mol-1 respectively. Then for (Λ0) NaBr is

(Λ0) NaBr =126+152 -150 0 =128S cm2 mol-1

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The limiting molar conductivities $(Λ^{0})$ for NaCl, KBr and KCl are 126, 152 and150 Scm²mol^-1 respectively. Then for $(Λ^{0})$ NaBr is

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answer is 128.

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$(Λ^{0})$ NaBr =126+152 -150 0 =128S cm² mol^-1

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