Question

Limiting molar conductivity of NaBr is

• A. $\Delta º_{m}NaBr = \Delta º_{m}NaCl + \Delta º_{m}KBr$
• B. $\Delta º_{m}NaBr = \Delta º_{m}NaCl + \Delta º_{m}KBr - \Delta_{m}^{0}KCl$
• C. $\Delta º_{m}NaBr = \Delta º_{m}NaOH + \Delta º_{m}NaBr - \Delta_{m}^{0}NaCl$
• D. $\Delta º_{m}NaBr = \Delta º_{m}NaCl - \Delta º_{m}KBr$

Answer 1

Answer: (B)

$\Delta º_{m}NaBr = \Delta º_{m}NaCl + \Delta º_{m}KBr - \Delta_{m}^{0}KCl$

Answer 2

$\Delta º_{m}NaBr = \Delta º_{m}NaCl + \Delta º_{m}KBr - \Delta º_{m}KCl$