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Question

Mohrs salt is

$N{{a}_{2}}S{{O}_{4}}.A{{l}_{2}}(S{{O}_{4}}).24{{H}_{2}}O$

$CuS{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O$

$FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O$

${{K}_{2}}S{{O}_{4}}.F{{e}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O$

Answer

$FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O$

Explanation

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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