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Limiting molar conductivity of NaBr isA. `Lambda_m^@NaBr = Lambda_m^@NaCl + Lambda_m^@KBr`B. `Lambda_m^@NaBr = Lambda_m^@ NaCl + Lambda_m^@KBr - Lambda_m^@KCl`C. `Lambda_m^@NaBr=Lambda_m^@NaOH + Lambda_m^@NaBr - Lambda_m^@NaCl`D. `Lambda_m^@NaBr=Lambda_m^@NaCl-Lambda_m^@NaBr` |
Answer» Correct Answer - B | |