Question

Molar conductances of $BaC{l}_2,H_{2}S{O}_4$ and HCl at infinite dilutions are $x_1,x_2$ and $x_3$ respectively. Equivalent conductance of $BaS{O}_4$ at infinite dilution will be :

• A. $\frac{(x_1+x_2-x_3)}{2}$
• B. $x_1+x_2-2x_3$
• C. $\frac{(x_1-x_2-x_3)}{2}$
• D. $\frac{(x_1+x_2-2x_3)}{2}$

Answer 1

The correct option is D

$\frac{(x_1+x_2-2x_3)}{2}$

${\lambda }_m^{\infty }\left(BaS{O}_4\right)={\lambda }_{b{A}^{2+}}^{\infty }+{\lambda }_{s{o}_4^{2-}}^{\infty }={\lambda }_m^{\infty }\left(BaC{l}_2\right)+{\lambda }_m^{\infty }(H_{2}S{O}_4{)}^{-}2{\lambda }_{\left(HCl\right)}^{\infty }=x_1+x_2-2x_3;\lambda {\lambda }_e^{\circ }=\frac{1}{2}{\lambda }_m^{\infty }(BaS{O}_4)$