Answer:
The salt (X)
dissolves and its aqueous solution is alkaline to litmus, it must be a salt of
a strong base and a weak acid. On heating it forms a glassy material, hence (X)
is a borax.
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}+7{{H}_{2}}O\xrightarrow{{}}\underset{Strong\,alkali}{\mathop{2NaOH}}\,+4{{H}_{3}}B{{O}_{3}}\]
\[\underset{(X)}{\mathop{N{{a}_{2}}{{B}_{4}}{{O}_{7}}}}\,\cdot
10{{H}_{2}}O\xrightarrow{Heat}N{{a}_{2}}{{B}_{4}}{{O}_{7}}+10{{H}_{2}}O\]
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}\xrightarrow{Heat}2NaB{{O}_{2}}+{{B}_{2}}{{O}_{3}}\]
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}\underbrace{N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}{{B}_{4}}{{O}_{7}}}_{\underset{(Y)}{\mathop{Glass\,material}}\,}\]
\[{{H}_{2}}{{B}_{4}}{{O}_{7}}+5{{H}_{2}}O\xrightarrow{{}}\underset{\underset{\begin{smallmatrix}
Orthoboric\,acid \\
(White\,crystals)
\end{smallmatrix}}{\mathop{(Z)}}\,}{\mathop{4{{H}_{3}}N{{O}_{3}}}}\,\]
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