Question

Find the value of $\Lambda_{ eq }^{\circ}$ for potash alum.
$\lambda^{\circ}_{m}\left(K^{+}\right) =73.5 \Omega^{-1}\,cm^{2}\, mol\,{}^{-1}$
$\lambda^{\circ}_{m} \left(Al^{3+}\right) = 189 \Omega^{-1}\, cm^{2}\, mol^{-1}$
$\lambda^{\circ}_{m} \left(SO^{2-}_{4}\right) = 160\Omega^{-1} cm^{2}\, mol^{-1}$

• A. $145.6 \Omega^{-1}\, cm^{2}\, eq^{-1}$
• B. $1165 \Omega^{-1}cm^{2} eq^{-1}$
• C. $532 \Omega^{-1} cm^{2} eq^{-1}$
• D. $195.5 \Omega^{-1}cm^{2} eq^{-1}$

Answer 1

Answer: (A)

$K_{2}SO_{4} \cdot Al_{2} \left(SO_{4}\right)_{3} \cdot 24H_{2}O$
$\Rightarrow 2K^{+}_{\left(aq\right)} +2Al^{3+}_{\left(aq\right)} +4SO^{2-}_{4\left(aq\right)}$
$\wedge^{\circ}_{m}$ (potash alum) $= 2\lambda^{\circ}_{m} \left(K^{+}\right) +2\lambda^{\circ}_{m} \left(Al^{3+}\right) +4\lambda^{\circ}_{m} \left(SO^{2-}_{4}\right)$
$= 2 \times73.5 +2 \times189 +4 \times160$
$= 1165 m^{-1}\, cm^{2}\, mol^{-1}$
$n$ for potash alum $= 8$ (total +ve charge)
$\wedge^{\circ}_{eq} = \frac{\wedge^{\circ}_{m}}{n} = \frac{1165}{8} = 145.6\Omega^{-1}\, cm^{2}\, eq^{-1}$