Some Basic Concepts of Chemistry - Result Question 40
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 $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$. 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
$A+\mathrm{S} \longrightarrow \underset{B}{M_2 \mathrm{SO}_4}$
$\because \quad 0.321 \mathrm{~g}$ sulphur gives 1.743 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
$2 \mathrm{KO}_x+\mathrm{S} \longrightarrow \mathrm{K}_2 \mathrm{SO}_4$
$\Rightarrow \quad x=2$, i.e. $A$ is $\mathrm{KO}_2$.
BETA
