A) Borax
B) Alum
C) Microcosmic salt
D) None
Correct Answer: B
Solution :
\[{{K}_{2}}S{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\xrightarrow[swells]{\Delta }\]\[\underbrace{{{K}_{2}}S{{O}_{4}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}_{Amorphous\,powder}+24{{H}_{2}}O\] \[{{K}_{2}}S{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O+8NaOH\xrightarrow[{}]{{}}\] \[\underset{so\operatorname{lub}le}{\mathop{2KOH+}}\,\underset{(white\,ppt.)}{\mathop{2Al{{(OH)}_{3}}}}\,\downarrow \underset{so\operatorname{lub}le}{\mathop{+4N{{a}_{2}}S{{O}_{4}}}}\,\]\[\xrightarrow[{}]{\begin{smallmatrix} Excess\,of \\ NaOH \end{smallmatrix}}\underset{so\operatorname{lub}le}{\mathop{{{[Al{{(OH)}_{4}}]}^{-}}}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
