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A colourless crystalline salt $X$ is soluble in dilute $\mathrm{HCl}$. On adding $\mathrm{NaOH}$ solution, it gives a white precipitate which is insoluble in excess of $\mathrm{NaOH}.X$ is
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The correct answer is:
$\mathrm{MgSO}_{4}$
$\mathrm{MgSO}_{4}$ is soluble in $\mathrm{HCl}$ and gives $\mathrm{MgCl}_{2}$ and
$\mathrm{MgSO}_{2} \mathrm{SO}_{4}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)$
$\mathrm{On}$ adding $\mathrm{NaOH}$ to $\mathrm{MgCl}_{2}$, it gives white ppt. of $\mathrm{Mg}(\mathrm{OH})_{2}$.
$\mathrm{MgCl}_{2}(a q)+\underset{\text { Excess }}{2 \mathrm{NaOH}(a q)} \longrightarrow \underset{\text { White ppt. }}{\mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{NaCl}}$
$\mathrm{MgSO}_{2} \mathrm{SO}_{4}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)$
$\mathrm{On}$ adding $\mathrm{NaOH}$ to $\mathrm{MgCl}_{2}$, it gives white ppt. of $\mathrm{Mg}(\mathrm{OH})_{2}$.
$\mathrm{MgCl}_{2}(a q)+\underset{\text { Excess }}{2 \mathrm{NaOH}(a q)} \longrightarrow \underset{\text { White ppt. }}{\mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{NaCl}}$
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