A) \[A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\]
B) \[NaCl\]
C) \[N{{a}_{2}}S{{O}_{4}}\]
D) \[Al{{(N{{O}_{3}})}_{3}}\]
Correct Answer: A
Solution :
\[{{K}_{4}}[Fe{{(CN)}_{6}}]\]dissociates as \[4{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{4-}},\] thus 1 molecule dissociates into five particles in the similar way \[A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\] also gives five particles per molecule.
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