Question

The molar conductance at infinite dilution of $BaC{l}_2$, $NaCl$, $NaOH$ are
$280\times {10}^{-4}$,$126.5\times {10}^{-4}$,$248\times {10}^{-4}S{m}^{2}mo{l}^{-1}$ respectively. The molar conductance at infinite dilution for $Ba(OH{)}_2$ is:

• A. $523\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• B. $52.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• C. $5.23\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• D. $65\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is A $523\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductance of the anions and cations.
Example for $A_n{B}_m$
${\lambda }^0=n{\lambda }_{A^{+}}^0+m{\lambda }_{B^{-}}^0$
So for given question:
${\lambda }_{BaC{l}_2}^0={\lambda }_{B{a}^{+}}^0+2{\lambda }_{C{l}^{-}}^0=280\times {10}^{-4}\left(i\right)$${\lambda }_{NaCl}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{C{l}^{-}}^0=126.5\times {10}^{-4}\left(ii\right)$
${\lambda }_{NaOH}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{O{H}^{-}}^0=248\times {10}^{-4}\left(iii\right)$
For
${\lambda }_{Ba(OH{)}_2}^0={\lambda }_{B{a}^{+}}^0+2{\lambda }_{O{H}^{-}}^0=i+2\left(iii\right)-2\left(ii\right)$
$=523\times {10}^{-4}S{m}^{2}mo{l}^{-1}$