Alkanline earth metal nitrate `(A)` on heating decompose, leaving a solid residue `(B)` which goes into solution with dilute `HCI`. The solution of `(B)` gives a white precipitate `(C )` is dissolved in dilute `HCl` and the solution is treated with potassium chromae to get yellow precipitate `(D)`. The solution `(B)` with dilute `H_(2)SO_(4)` also gives a white precipitate `(E)` insoluble in diute `HCl` and nitric acid. the precipitate `(E)` is a part of a white pigment lithopone.
The compound `(E)` is

Options

A. `BaSO_(4)`
B. `MgSO_(4)`
C. `CaSO_(4)`
D. `Na_(2)SO_(4)`

Answer

A

Explanation

White ppt. `(E)` is a part of white pigment lithopone. Lithopone is a mixture of `BaSO_(4)` and `ZnS. BaSO_(4)` is white ppt. which is insoluble in dil `HCl` and `HNO_(3)`, whereas `ZnS` is soluble.
`ZnS+2HClrarrZnCl_(2)+H_(2)S`
`ZnS+2HNO_(3)rarrZn(NO_(3))_(2)+H_(2)S`
Hence, `(E)` is `BaSO_(4)`.
Alkaline earth metal nitrate, `(A)` is `Ba (NO_(3))_(2)`.
`underset((A))(Ba(NO_(3))_(2)overset(Delta) underset("Solid residue")underset((B))(BaO) +2NO_(2)+(1)/(2)O_(2)`
`underset((B))(BaO)+underset(dil)(2HCI)rarrunderset("in dil HCI")underset("Soluble")(BaO) +H_(2)O`
`BaCI_(2) +underset(dil)(H_(2)SO_(4) rarr underset((E))underset("White ppt.")(BaSO_(4))+2HCI`
`BaCI_(2) +(NH_(4))_(2) CO_(3) rarr underset((C))underset("White ppt.")(BaCO_(3))+2NH_(4)CI`
`BaCO_(3) +underset(dil)(2HCI) rarr underset("soluble")(BaCI_(2) +CO_(2) +H_(2)O`
`BaCI_(2) +underset(underset("chromate")("Potassium"))(K_(2)CrO_(4))rarr underset((D))(underset("precipitate")underset(Yellow)(Ba(CrO_(4))))+2KCI`
`(E)` is `BaSO_(4)`