Question

Given the ionic equivalent conductivities for the following ions :

${\lambda }_{eq}^{\circ }{K}^{\oplus }=73.5c{m}^{2}oh{m}^{-1}e{q}^{-1}$

${\lambda }_{eq}^{\circ }A{l}^{3+}=189c{m}^{2}oh{m}^{-1}e{q}^{-1}$

${\lambda }_{eq}^{\circ }S{O}_4^{2-}=85.8c{m}^{2}oh{m}^{-1}e{q}^{-1}$

The ${\Lambda }_{eq}^{\circ }$ for potash alum $(K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O)$ is :

• A. $245.92$
• B. $348.3$
• C. $368.2$
• D. $108.52$

Answer 1

Answer: (A)

$245.92$

(Eq = Charge on the ion/Total charge)

$\left[K^{\oplus }\right]=\frac{1}{8}mol\times 2=\frac{1}{4}mol=\frac{1}{4}Eq$

$\left[A{l}^{3+}\right]=\frac{6}{8}=\frac{3}{4}Eq$

$\left[S{O}_4^{-2}\right]=\frac{8}{8}=1Eq$

Now,

${\Lambda }_{eq}^{\circ }{K}_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O$

$={\lambda }_{eq}^{\circ }\left(K^{\oplus }\right)+{\lambda }_{eq}^{\circ }\left(A{l}^{3+}\right)+{\lambda }_{eq}^{\circ }\left(S{O}_4^{2-}\right)$

$\equiv \frac{1}{4}\times 73.5+189\times \frac{3}{4}+85.8\times 1$

$\equiv 18.375+141.75+85.8$

$\equiv 245.92$