Question

The limiting molar conductivities $\Lambda^{0}$ for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol–1 respectively. The value of $\Lambda^{0}$ for NaBr is

• A. 128 S cm2mol–1
• B. 176 S cm2mol–1
• C. 278 S cm2mol–1
• D. 302 S cm2mol–1

Answer 1

Answer: (A)

128 S cm2mol–1

Answer 2

${{}_{\land}^{0}}_{NaBr}$= 126 + 152 – 150 = 128 S cm2 mol–1.