A) \[0.1N\,N{{a}_{2}}S{{O}_{4}}\]
B) \[0.1N\,MgS{{O}_{4}}\]
C) \[0.1M\,A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\]
D) \[0.1M\,BaS{{O}_{4}}\]
Correct Answer: C
Solution :
\[A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\] gives maximum ion hence it will show highest boiling point.
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