A) Borax
B) Alum
C) Microcosmic salt
D) None
Correct Answer: B
Solution :
\[{{K}_{2}}S{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\xrightarrow[swells]{\Delta }\underbrace{{{K}_{2}}S{{O}_{4}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}_{Amorphous\,powder}+24{{H}_{2}}O\] \[{{K}_{2}}S{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O+8NaOH\xrightarrow{{}}\underset{\text{soluble}}{\mathop{2KOH}}\,+\underset{(white\,ppt.)}{\mathop{2Al{{(OH)}_{3}}}}\,\]\[\downarrow \underset{\text{soluble}}{\mathop{+4N{{a}_{2}}S{{O}_{4}}}}\,\xrightarrow{\begin{smallmatrix} Excess\,\,of \\ NaOH \end{smallmatrix}}\underset{\text{soluble}}{\mathop{{{[Al{{(OH)}_{4}}]}^{-}}}}\,\]
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