A) \[N{{a}_{2}}S{{O}_{4}}.A{{l}_{2}}(S{{O}_{4}}).24{{H}_{2}}O\]
B) \[CuS{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\]
C) \[FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O\]
D) \[{{K}_{2}}S{{O}_{4}}.F{{e}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\]
Correct Answer: C
Solution :
(i) \[N{{a}_{2}}S{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\]and \[{{K}_{2}}S{{O}_{4}}.F{{e}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\]are alum. (ii) \[FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O\]is Mohrs salt.
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