Question

From the following molar conductivities at infinite dilution,

${\Lambda }_m^o\text{for}A{l}_2{\left(S{O}_4\right)}_3=858Sc{m}^{2}mo{l}^{-1}$
${\Lambda }_m^o\text{for}N{H}_{4}OH=238.3Sc{m}^{2}mo{l}^{-1}$
${\Lambda }_m^o\text{for}{\left(N{H}_4\right)}_{2}S{O}_4=238.4Sc{m}^{2}mo{l}^{-1}$

calculate ${\Lambda }_m^o\text{for}Al{\left(OH\right)}_3$.

• A. 715.2 $Sc{m}^{2}mo{l}^{-1}$
• B. 1575.6 $Sc{m}^{2}mo{l}^{-1}$
• C. 786.3 $Sc{m}^{2}mo{l}^{-1}$
• D. 157.56 $Sc{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (C)

786.3 $Sc{m}^{2}mo{l}^{-1}$

Molar conductivities at infinite dilution :

${\Lambda }_m^0$ for ${Al}_2{\left({SO}_4\right)}_3=858$ ${Scm}^2{mol}^{-1}$

${\Lambda }_m^0$ for ${NH}_{4}OH=238.3$ ${Scm}^2{mol}^{-1}$

${\Lambda }_m^0$ for ${\left({NH}_4\right)}_2{SO}_4=238.4$ ${Scm}^2{mol}^{-1}$

${Al}_2{\left({SO}_4\right)}_3\rightleftharpoons {2Al}^{3+}+{3SO}_4^{2-}$

${NH}_{4}OH\rightleftharpoons {NH}_4^{+}+{OH}^{-}$

${\left({NH}_4\right)}_2{SO}_4\rightleftharpoons 2{NH}_4^{+}+{SO}_4^{2-}$

${Al\left(OH\right)}_3\rightleftharpoons {Al}^{3+}+{3OH}^{-}$

$2\left({\Lambda }_m^0{Al\left(OH\right)}_3\right)={\Lambda }_m^0{Al}_2{\left({SO}_4\right)}_3+6\left({\Lambda }_m^0{NH}_{4}OH\right)-3\left({\Lambda }_m^0{\left({NH}_4\right)}_2{SO}_4\right)$

$2\left({\Lambda }_m^0{Al\left(OH\right)}_3\right)=858+6\left(238.3\right)-3\left(238.4\right)$

$=1572.6$

${\Lambda }_m^0{Al\left(OH\right)}_3=\frac{1572.6}{2}=786.3$ ${Scm}^2{mol}^{-1}$

Hence, option C is correct.