A) \[{{K}_{2}}S{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\]
B) \[{{K}_{4}}[Fe{{(CN)}_{6}}]\]
C) \[{{K}_{2}}S{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.6{{H}_{2}}O\]
D) \[N{{a}_{2}}C{{O}_{3}}.10{{H}_{2}}O\]
Correct Answer: A
Solution :
General formula for alum is \[{{M}_{2}}S{{O}_{4}}.{{R}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\] \[M=\] mono valent cation \[({{K}^{+}},\,N{{a}^{+}}....)\] \[R=\]Trivalent cation \[(A{{l}^{+3}},\,F{{e}^{+3}})\] Hence, \[{{K}_{2}}S{{O}_{4}}A{{l}_{2}}{{(S{{O}_{4}})}_{2}}.24{{H}_{2}}O\] represent an alum.
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