Some Basic Concepts of Chemistry - Result Question 110
####48. $A$ 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 hydrated double salt, $C$ with $Al _2\left(SO _4\right) _3$. Identify $A, B$ and $C$.
(1994, 2M)
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Solution:
- Compound $B$ forms hydrated crystals with $Al _2\left(SO _4\right) _3$. 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
$$ A+S \longrightarrow M _2 SO _4 $$
$\because \quad 0.321 g$ sulphur gives $1.743 g$ of $M _2 SO _4$
$\therefore \quad 32.1 g S$ (one mole) will give $174.3 g M _2 SO _4$
Therefore, molar mass of $M _2 SO _4=174.3 g$
$\Rightarrow \quad 174.3=2 \times$ Atomic weight of $M+32.1+64$
$\Rightarrow$ Atomic weight of $M=39$, metal is potassium (K)
$K _2 SO _4$ on treatment with aqueous $Al _2\left(SO _4\right) _3$ gives potash-alum.
$K _2 SO _4+Al _2\left(SO _4\right) _3+24 H _2 O \longrightarrow K _2 SO _4 Al _2\left(SO _4\right) _3 \cdot 24 H _2 O$
If the metal oxide $A$ has molecular formula $M O _x$, fwo moles of it combine with one mole of sulphur to give one mole of metal sulphate as
$$ \begin{aligned} & 2 KO _x+S \longrightarrow K _2 SO _4 \ & \Rightarrow \quad x=2 \text {, i.e. } A \text { is } KO _2 . \end{aligned} $$
