Question

A mixture of two white substances is soluble in water. This solution gives brown colour gas on passing chlorine gas. Another sample of solution gives white precipitate with $BaC{l}_2$ which is insoluble in concentrated $HCl$. The original solution of the mixture gives white precipitate with large excess of $NaOH$ solution whose suspension is used as antacid. After filtering the precipitate, the filtrate was boiled with excess $NaOH$ solution which gave a yellowish precipitate on adding $NaCl{O}_4$. One of the compound of the mixture forms alum. Identify the mixture.

• A. Mixture consists $K_{2}S{O}_4$ and $MgB{r}_2$
• B. Mixture consists $K_{2}S{O}_3$ and $MgBr$
• C. Mixture consists $K_{2}S{O}_4$ and $MgBr$
• D. None of these

Answer 1

The correct option is A Mixture consists $K_{2}S{O}_4$ and $MgB{r}_2$

The mixture consists of $K_{2}S{O}_4$ and $MgB{r}_2$.

Both are white substances soluble in water. This solution gives brown colour gas on passing chlorine gas. This indicates the replacement of bromine with chlorine.
$MgB{r}_2+C{l}_2\to MgC{l}_2+B{r}_2$

Another sample of solution gives white precipitate with $BaC{l}_2$ which is insoluble in concentrated $HCl$. Barium chloride reacts with potassium sulphate to form white ppt of barium sulphate.
$BaC{l}_2+K_{2}S{O}_4\to 2KCl+BaS{O}_4\downarrow \text{(white)}$

The original solution of the mixture gives white precipitate of magnesium hydroxide with large excess of NaOH solution whose suspension is used as an antacid.
$MgB{r}_2+2NaOH\to Mg(OH{)}_2\downarrow \text{(white ppt)}+2NaBr$

After filtering of this precipitate, the filtrate was boiled with excess NaOH solution gave a yellowish precipitate on adding $NaCl{O}_4$.
$K_{2}S{O}_4+2NaCl{O}_4\to N{a}_{2}S{O}_4+2KCl{O}_4$

One of the compounds of the mixture forms alum. The specific compound is the hydrated potassium aluminium sulfate (potassium alum) with the formula $KAl(S{O}_4{)}_2\cdot 12H_{2}O$.