Question

(A) is binary orange coloured paramnetic compound of a univalent cation. 1.422 g of (A) reacts completely with 0.321 g of sulphur ina evacuated and sealed tube to give 1.743 g of a while crystalline solid (B) that forms hydrated double salt (C) with $A{l}_2(S{O}_4{)}_3$; Here, compounds (A), (B), and (C) are $K{O}_2$, $K_{2}S{O}_4$ and $K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O$ respectively.
If true enter 1, else enter 0.

Answer 1

(C) is hydrated double salt formed from (B) and $A{l}_2(S{O}_4{)}_3$
$\Rightarrow$ (C) can be an Alum
$M_{2}S{O}_4.$ $A{l}_2(S{O}_4{)}_3$ $x{H}_{2}O$
(B) is obtained when S reacts with (A) is a sealed tube
$\Rightarrow$ (A) is also a source of oxygen that can convert S into $S{O}_4^{2-}$
(A) has univalent cation $M^{+}$ and its compound with oxygen can be
$M_{2}O$ (oxide)
$M_2{O}_2$ (peroxide)
$M{O}_2$ (superoxide)
Only $M{O}_2$ is paramagnetic
$\Rightarrow$ A is $M{O}_2$ (binary given)
$\underset{2(M+32)}{2M{O}_2}+\underset{32}{S}⟶\underset{(2M+96)}{M_{2}S{O}_4}$
$2(M+32)gM{O}_2$ reacts with 32 of sulphur
$\therefore$ 1.422 g $M{O}_2$ reacts with $\frac{32}{2(M+32)}x1.422g$ sulphur
$\frac{32X1.422}{2(M+32)}$ = 0.321 g sulphur
$\Rightarrow$ M = 39 (K)
$\Rightarrow$ (A) is $K{O}_2$
(B) is $K_{2}S{O}_4$
(C) is $K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O$ (potash alum)
[value of M can also be evaluated by using weight of $M_{2}S{O}_4$(i.e. 1.743 g (given))]
$\frac{weightofS}{weightof{M}_{2}S{O}_4}$ = $\frac{32}{(2M+96)}$
$\frac{0.321}{1.743}$= $\frac{32}{2M+96}$