Question

Given: ${\Lambda }^{\infty }\left(\frac{1}{3}A{l}^{3+}\right)=63{\Omega }^{-1}c{m}^{2}e{q}^{-1}$ and ${\Lambda }^{\infty }\left(\frac{1}{2}S{O}_4^{2-}\right)=80{\Omega }^{-1}c{m}^{2}e{q}^{-1}$. Then the value ${\Lambda }^{\infty }\left(A{l}_2\right(S{O}_4{)}_3)$ would be:

• A. $143{\Omega }^{-1}c{m}^{2}e{q}^{-1}$
• B. $858{\Omega }^{-1}c{m}^{2}e{q}^{-1}$
• C. $206{\Omega }^{-1}c{m}^{2}e{q}^{-1}$
• D. $286{\Omega }^{-1}c{m}^{2}e{q}^{-1}$

Answer 1

The correct option is A $143{\Omega }^{-1}c{m}^{2}e{q}^{-1}$

According to kohlrausch's law of independent migration of ions, the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of contributions of the molar conductivities of its ions.
Hence,
${\Lambda }^{\infty }\left({Al}_2{\left({SO}_4\right)}_3\right)={\lambda }^{\infty }\left(\frac{1}{3}{Al}^{3+}\right)+{\lambda }^{\infty }\left(\frac{1}{2}{SO}_4^{2-}\right)$
Substitute the values in the above equation,
${\Lambda }^{\infty }\left({Al}_2{\left({SO}_4\right)}_3\right)=63+80=143{\Omega }^{-1}{cm}^2{eq}^{-1}$

Hence, option A is correct.