Question

Molar conductances of $BaCl_{2}, H_{2}SO_{4}$ and $HCl$ at infinite dilutions are $x_{1}, x_{2}$ and $x_{3}$, respectively. Equivalent conductance of $BaSO_{4}$ at infinite dilution will be:

• A. $\frac{[x_{1}+x_{2} -x_{3} ]}{2}$
• B. $\frac{[x_{1}-x_{2} -x_{3} ]}{2}$
• C. $2(x_{1} +x_{2} -2x_{3})$
• D. $\frac{[x_{1}+x_{2} -2x_{3} ]}{2}$

Answer 1

Answer: (D)

$\Lambda_{m,BaCl_2} =\Lambda_{m Ba^{2+}}+2\Lambda_{m\, cl^{-}}\,\, x_{1}$
$\Lambda_{m H_2SO_4} =2\Lambda_{mH^{+}}+ \Lambda_{m\,SO^{-}_{4} }\,\, x_{2}$
$\Lambda_{m HCl} =\Lambda_{mH^{+}} +\Lambda_{m\,Cl^{-}} \,\, x_{3}] \times2$
$\therefore \Lambda_{m, BaSO_4} =\left(x_{1}+x_{2}-2x_{3}\right)$
$\:\Rightarrow \:\Lambda _{eq}.\:BaSO_4\:=\frac{\Lambda _{m,\:BaSO_4}}{\text{Total charge (cation/anion)}}$
$\:\Lambda _{eq,\:BaSO_4}\:=\frac{\left[x_1\:+x_2\:-2x_3\right]}{2}$