A is a binary compound of a univalent metal. 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743 g of a white crystalline solid B, that forms a bydrated double salt, C with Al_(2) (SO_(4))_(3). Identify A, B and C.

A is a binary compound of a univalent metal. 1.422 g of A reacts compl

1.	A is a binary compound of a univalent metal. 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube to give 1.743 g of a white crystalline solid B, that forms a bydrated double salt, C with Al_(2) (SO_(4))_(3). Identify A, B and C.
Answer» Solution :Compound B FORMS hydrated CRYSTALS with `Al_(2)(SO_(4))_(3)`. Also, B is formed with univalent metal on heating with sulphur. Hence, compound B MUST has the molecular formula `M_(2)SO_(4)` and compound A must be an oxide of M which reacts with sulphur to give metal sulphate as `A+S rarr M_(2)SO_(4)` `:' 0.321 g` sulphur gives `1.743 g` of `M_(2)SO_(4)` `:. 32.1 g S` (one MOLE) will give `174.3 g M_(2)SO_(4)` Therefore, molar mass of `M_(2)SO_(4)=174.3 g` `implies 174.3=2xx` Atomic weight of `M+32.1+64` `implies` Atomic weight of `M=39`, metal is potassium `(K)` `K_(2)SO_(4)` on treatement with aqueous `Al_(2)(SO_(4))_(3)` gives potash-alum. `underset(B)(K_(2)SO_(4))+Al_(2)(SO_(4))_(3)+24H_(2)O rarr K_(2)SO_(4)Al_(2)underset(C)((SO_(4))_(3)).24H_(2)O` If the metal oxide A has molecular formula `MO_(x)` two moles of it combine with one mole of sulphur to give one mole of metal sulphate as `2KO_(x)+S rarr K_(2)SO_(4)` `implies x=2`, i.e. `A` is `KO_(2)`.