Question

A solution of a compound X in dilute HCl on treatment with a solution of $BaCl_{2}$gives a white precipitate of a compound Y which is insoluble in conc. $HNO_{3}$and conc. HCl. Compound X imparts golden yellow colour to the flame.

\mathbf{\& IndiluteHCl)} \\ \mathbf{\&(impartsgolden} \\ \mathbf{\& Yellowcolour)} \end{array}}{\mathbf{(Solution}}\mathbf{+ BaC}\mathbf{l}_{\mathbf{2}}\mathbf{\rightarrow}\underset{\begin{array}{r} \mathbf{\& W}\mathbf{h}\mathbf{ite} \\ \mathbf{\& ppt} \end{array}}{\mathbf{Y}}\underset{\mathbf{ConcHCl}}{\overset{\mathbf{\quad}\overset{\mathbf{Conc.HN}\mathbf{o}_{\mathbf{3}}}{\mathbf{+}}\mathbf{\quad}}{\rightarrow}}\mathbf{Inso}\mathbf{lub}\mathbf{l}\mathbf{e}$$ What are compounds X and Y?

• A. X is $MgCl_{2}$and Y is $BaSO_{4}$
• B. X is $CaCl_{2}$and Y is $BaSO_{4}$
• C. X is $Na_{2}SO_{4}$ and Y is $BaSO_{4}$
• D. X is $MgSO_{4}$and Y is $BaSO_{4}$

Answer 1

Answer: (C)

X is $Na_{2}SO_{4}$ and Y is $BaSO_{4}$

Answer 2

: $\underset{(X)}{Na_{2}}SO_{4} + BaCl_{2} \rightarrow \underset{(Y)}{Ba}SO_{4} + 2NaCl$