Some Basic Concepts of Chemistry 1 Question 27
48. $A$ is a binary compound of a univalent metal. $1.422 \mathrm{~g}$ of $A$ reacts completely with $0.321 \mathrm{~g}$ of sulphur in an evacuated and sealed tube to give $1.743 \mathrm{~g}$ of a white crystalline solid $B$, that forms a hydrated double salt, $C$ with $\mathrm{Al}{2}\left(\mathrm{SO}{4}\right)_{3}$. Identify $A, B$ and $C$.
(1994, 2M)
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Solution:
- Compound $B$ forms hydrated crystals with $\mathrm{Al}{2}\left(\mathrm{SO}{4}\right){3}$. Also, $B$ is formed with univalent metal on heating with sulphur. Hence, compound $B$ must has the molecular formula $M{2} \mathrm{SO}_{4}$ and compound $A$ must be an oxide of $M$ which reacts with sulphur to give metal sulphate as
$$ A+\mathrm{S} \longrightarrow M_{2} \mathrm{SO}_{4} $$
$\because \quad 0.321 \mathrm{~g}$ sulphur gives $1.743 \mathrm{~g}$ of $M_{2} \mathrm{SO}_{4}$
$\therefore \quad 32.1 \mathrm{~g} \mathrm{~S}$ (one mole) will give $174.3 \mathrm{~g} M_{2} \mathrm{SO}_{4}$
Therefore, molar mass of $M_{2} \mathrm{SO}_{4}=174.3 \mathrm{~g}$
$\Rightarrow \quad 174.3=2 \times$ Atomic weight of $M+32.1+64$
$\Rightarrow$ Atomic weight of $M=39$, metal is potassium (K)
$\mathrm{K}{2} \mathrm{SO}{4}$ on treatment with aqueous $\mathrm{Al}{2}\left(\mathrm{SO}{4}\right)_{3}$ gives potash-alum.
$\mathrm{K}{2} \mathrm{SO}{4}+\mathrm{Al}{2}\left(\mathrm{SO}{4}\right){3}+24 \mathrm{H}{2} \mathrm{O} \longrightarrow \mathrm{K}{2} \mathrm{SO}{4} \mathrm{Al}{2}\left(\mathrm{SO}{4}\right){3} \cdot 24 \mathrm{H}{2} \mathrm{O}$
If the metal oxide $A$ has molecular formula $M \mathrm{O}_{x}$, fwo moles of it combine with one mole of sulphur to give one mole of metal sulphate as
$$ \begin{aligned} & 2 \mathrm{KO}{x}+\mathrm{S} \longrightarrow \mathrm{K}{2} \mathrm{SO}{4} \ & \Rightarrow \quad x=2 \text {, i.e. } A \text { is } \mathrm{KO}{2} . \end{aligned} $$
