MCQ
The formula of alum is
- ✓${K_2}S{O_4}.A{l_2}{(S{O_4})_3}.24{H_2}O$
- B${K_4}[Fe{(CN)_6}]$
- C${K_2}S{O_4}.A{l_2}{(S{O_4})_3}.6{H_2}O$
- D$N{a_2}C{O_3}.10{H_2}O$
${M_2}S{O_4}.{R_2}{(S{O_4})_3}.24{H_2}O$
$M = $ mono valent cation $({K^ + },\,N{a^ + }....)$
$R = $Trivalent cation $(A{l^{ + 3}},\,F{e^{ + 3}})$
Hence, ${K_2}S{O_4}A{l_2}{(S{O_4})_2}.24{H_2}O$ represent an alum.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$\begin{array}{*{20}{c}}
H\\
|\\
{ClHC = HC - C - CH = CHCl}\\
|\\
{Cl}
\end{array}$
$I = -3.04\,V$,
$II = -1.90\,V$,
$III = 0\,V$,
$IV = 1.90\,V$