Question

$A_M^{\infty }$ and $A_{eq.}^{\infty }$ are molar and equivalent conductivities at infinite dilution; $\lambda$ is ionic conductivity at infinite dilution, then for potash alum:

• A. $A_M^{\infty }$ (potash alum)$=2\times {\lambda }_{K^{+}}^{\infty }+2\times {\lambda }_{A{l}^{3+}}^{\infty }+4\times {\lambda }_{S{O}_4^{2-}}^{\infty }$
• B. $A_e^{\infty }q$ (potash alum)$=\frac{1}{8}\times A_M^{\infty }$ (potash alum)
• C. $A_M^{\infty }$ (potash alum)$=\frac{1}{4}\times {\lambda }_{K^{+}}^{\infty }+\frac{1}{4}\times {\lambda }_{A{l}^{3+}}^{\infty }+\frac{1}{2}\times {\lambda }_{S{O}_4^{2-}}^{\infty }$
• D. $A_M^{\infty }$ (potash alum)$={\lambda }_{K^{+}}^{\infty }+{\lambda }_{A{l}^{3+}}^{\infty }+{\lambda }_{S{O}_4^{2-}}^{\infty }$

Answer 1

Answer: (A), (B)

$A_M^{\infty }$ (potash alum)$=2\times {\lambda }_{K^{+}}^{\infty }+2\times {\lambda }_{A{l}^{3+}}^{\infty }+4\times {\lambda }_{S{O}_4^{2-}}^{\infty }$
$A_e^{\infty }q$ (potash alum)$=\frac{1}{8}\times A_M^{\infty }$ (potash alum)

$A_M^{\infty }$ and $A_{eq.}^{\infty }$ are molar and equivalent conductivities at infinite dilution; $\lambda$ is ionic conductivity at infinite dilution, then for potash alum

(A) $A_M^{\infty }$ (potash alum)$=2\times {\lambda }_{K^{+}}^{\infty }+2\times {\lambda }_{A{l}^{3+}}^{\infty }+4\times {\lambda }_{S{O}_4^{2-}}^{\infty }$

The molar conductivity is equal to the sum of the ionic conductivities.

(B) $A_{eq}^{\infty }$ (potash alum)$=\frac{1}{8}\times A_M^{\infty }$ (potash alum) [Note : charge will be 8]

The equivalent conductivity is one eight of the molar conductivity.