A) \[\text{2}{{\Lambda }^{\text{o}}}_{A{{l}^{3+}}}+3{{\Lambda }^{\text{o}}}_{so\frac{2-}{4}}\]
B) \[{{\Lambda }^{\text{o}}}_{A{{l}^{3+}}}+{{\Lambda }^{\text{o}}}_{so\frac{2-}{4}}\]
C) \[\text{(}{{\Lambda }^{\text{o}}}_{A{{l}^{3+}}}+3{{\Lambda }^{\text{o}}}_{so\frac{2-}{4}})\times 6\]
D) \[\frac{1}{3}{{\Lambda }^{\text{o}}}_{A{{l}^{3+}}}+\frac{1}{2}{{\Lambda }^{\text{o}}}_{so\frac{2-}{4}}\]
Correct Answer: B
Solution :
\[A{{l}_{2}}{{(S{{O}_{4}})}_{3}}2A{{l}^{3+}}+3SO_{4}^{2-}\]. Since equivalent conductance?s are given only for ions, the equivalent conductance at infinite dilution, \[\Lambda _{eq}^{\infty }=\Lambda _{Al\,3+}^{o}+\Lambda _{so_{4}^{2-}}^{o}\]
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