Question

According to Kohlrausch law, the limiting value of equivalent conductivity of an electrolyte $A_{2}B$ is given by:

• A. ${\lambda }_{A^{+}}^{\infty }+{\lambda }_{B^{2-}}^{\infty }$
• B. $1/2{\lambda }_{A^{+}}^{\infty }+{\lambda }_{B^{2-}}^{\infty }$
• C. ${\lambda }_{A^{+}}^{\infty }+1/2{\lambda }_{B^{2-}}^{\infty }$
• D. $2{\lambda }_{A^{+}}^{\infty }+{\lambda }_{B^{2-}}^{\infty }$

Answer 1

The correct option is D $2{\lambda }_{A^{+}}^{\infty }+{\lambda }_{B^{2-}}^{\infty }$

According to Kohlrausch law, the limiting value of equivalent conductivity of an electrolyte $A_{2}B$ is given by :

$2{\lambda }_{A^{+}}^{\infty }+{\lambda }_{B^{2-}}^{\infty }$.

since dissociation take place as shown

$A_{2}B⟶2A^{+}+B^{2-}$

It is the sum of the contributions from the cation and anion.