Question

A mixture of two white substances was dissolved in water. On passing $C{l}_2$ gas through the solution, a deep brown colour is developed. Addition of $BaC{l}_2$ solution to the original solution gives a white precipitate. Addition of a large amount of $NaOH$ solution to the original solution gives a white precipitate, whose suspension in water is used as an antacid. The mixture gives golden yellow colour flame. The mixture contains :

• A. $N{a}^{+},M{g}^{2+},B{r}^{-}$ and $S{O}_4^{2-}$
• B. $N{a}^{+},C{a}^{2+},B{r}^{-}$ and $S{O}_4^{2-}$
• C. $N{a}^{+},A{l}^{3+},B{r}^{-}$ and $S{O}_4^{2-}$
• D. $N{a}^{+},Z{n}^{2+},B{r}^{-}$ and $S{O}_4^{2-}$

Answer 1

The correct option is A $N{a}^{+},M{g}^{2+},B{r}^{-}$ and $S{O}_4^{2-}$

A deep brown colour of bromine gas is obtained when chlorine is passed through the given solution of the mixture.
$2B{r}^{-}+C{l}_2\to B{r}_2\uparrow \left(\text{reddish brown}\right)+2C{l}^{-}$
White precipitate of barium sulphate is formed after the addition of $BaC{l}_2$.
$S{O}_4^{2-}+BaC{l}_2\to BaS{O}_4\left(\text{white ppt.}\right)+2C{l}^{-}$
The white precipitate formed after the addition of $NaOH$ will be $Mg(OH{)}_2$ as it is used as an antacid.
$M{g}^{2+}+2NaOH\to Mg\left(OH{)}_2\right(\text{white ppt.})+2N{a}^{+}$
Golden yellow color of the flame shows that the mixture has sodium salt.
Thus, the mixture contains $N{a}^{+},M{g}^{2+},B{r}^{-}$ and $S{O}_4^{2-}$.