A) \[KMn{{(S{{O}_{4}})}_{2}}.12{{H}_{2}}O\]
B) \[KCr(S{{O}_{4}})2.12{{H}_{2}}O\]
C) \[N{{H}_{4}}Fe{{(S{{O}_{4}})}_{2}}.12{{H}_{2}}O\]
D) \[FeS{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\]
Correct Answer: D
Solution :
Pseudo alum has the general formula \[{{M}^{II}}S{{O}_{4}}.M_{2}^{III}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O.\] Therefore, \[FeS{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\]is a pseudo alum.
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