Question

Molar conductivity for a compound AB is $145.0Sc{m}^{2}mo{l}^{-1}$ and for CB is $110.1Sc{m}^{2}mo{l}^{-1}$. Limiting molar conductivity for $A^{+}$ is $73.5Sc{m}^{2}mo{l}^{-1}$. What is the molar conductivity for $C^{+}$?

• A. $326.6Sc{m}^{2}mo{l}^{-1}$
• B. $181.6Sc{m}^{2}mo{l}^{-1}$
• C. $38.6Sc{m}^{2}mo{l}^{-1}$
• D. $90.8Sc{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is C $38.6Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m\text{of AB}=145.05c{m}^{2}mo{l}^{-1}{\Lambda }_m\text{of AB}={\Lambda }_{A^{+}}^0+{\Lambda }_{B^{-}}^0\text{(According to Kohlraush's Law)}145={\Lambda }_{A^{+}}^0+{\Lambda }_{B^{-}}^0...\left(1\right){\Lambda }_m\text{of CB}=110.10Sc{m}^{2}mo{l}^{-1}{\Lambda }_m\text{of CB}={\Lambda }_{C^{+}}^0+{\Lambda }_{B^{-}}^{0}110.1={\Lambda }_{C^{+}}^0+{\Lambda }_{B^{-}}^0...\left(2\right)\text{Eqn(1) - Eqn(2)}$
$145-110.1={\Lambda }_{A^{+}}^0+{\Lambda }_{B^{-}}^0-{\Lambda }_{C^{+}}^0-{\Lambda }_{B^{-}}^{0}34.9={\Lambda }_{A^{+}}^0-{\Lambda }_{C^{+}}^{0}34.9=73.5-{\Lambda }_{C^{+}}^0{\Lambda }_{C^{+}}^0=73.5-34.9=38.6Sc{m}^{2}mo{l}^{-1}$