Chemistry - Electrochemistry Question with Solution | TestHub

${\Delta }_{{\mathrm{Na}}^{+}}^{\circ }+{\Delta }_{{\mathrm{OH}}^{-}}^{\circ }=248\times 10^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$ ${\Delta }_{{\mathrm{Na}}^{+}}^{\circ }+{\Delta }_{{\mathrm{Cl}}^{-}}^{\circ }=126\times 10^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$ ${\Delta }_{\mathrm{Ba}}^{\circ }{}^{2+}+\Delta {}_{2{\mathrm{Cl}}^{-}}^{\circ }=280\times 10^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$ $\mathrm{NoW}$, ${\Delta }_{\mathrm{Ba}(\mathrm{OH}{)}_2}^{\circ }={\Delta }_{{\mathrm{BaCl}}_2}^{\circ }+2\Delta {}^{\circ }{}_{\mathrm{NaOH}}^{\circ }-2\Delta {}_{\mathrm{NaC}1}^{\circ }$ $\Delta {}_{\mathrm{Ba}(\mathrm{OH}{)}_2}^{\circ }=280\times 10^{-4}+2\times 248\times 10^{-4}-2\times 126\times 10^{-4}$ ${\Delta }_{\mathrm{Ba}(\mathrm{OH}{)}_2}^{\circ }=524\times 10^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$