Question

In the aqueous solution, bleaching powder behaves as $CaOC{l}_2+H_{2}O\to C{l}_2+Ca(OH{)}_2$.

The equivalent weight of $CaOC{l}_2$ is $\frac{M}{x}$, where x is :

Answer 1

There are two types of Cl in $CaOC{l}_2$.

$C{l}^{-}$ (oxidation number = -1) oxidised to $C{l}_2$.

$Cl{O}^{-}$ (oxidation number = +1) reduced to $C{l}_2$.

Its a reverse of disproportionation reaction, when a compound in an intermediate oxidation state is formed from compounds in lower and higher oxidation states, is called comproportionation, also known as symproportionation.

$Ca\left(OCl\right)Cl=C{l}_2=2Cl$
+1 -1 0 0

$CaOC{l}_2=2Cl$

Here, one mole of $CaOC{l}_2$ reacts with 2 moles of electron.

$\therefore \text{Equivalent weight}=\frac{\text{Molecular weight}}{2}$
Hence the value of X is $2$