Question

To coagulate $\left[AgI\right]A{g}^{+}⋮N{O}_3^{-}SoI$; The decreasing order of coagulation power in following is -
$NaCI$
$K_{2}S{O}_4$
$C{a}_3(P{O}_4{)}_2$
$K_4\left[Fe\right(CN{)}_6]$

• A. $\left(a\right)>\left(b\right)>\left(c\right)>\left(d\right)$
• B. $\left(d\right)>\left(c\right)>\left(b\right)>\left(a\right)$
• C. $\left(a\right)>\left(d\right)>\left(c\right)>\left(b\right)$
• D. $\left(d\right)>\left(c\right)>\left(a\right)>\left(b\right)$

Answer 1

The correct option is B $\left(d\right)>\left(c\right)>\left(b\right)>\left(a\right)$

By Hardy Schulze rule coagulating power of an electrolyte is directly proportional to valency.

$NaCl$ $K_{2}S{O}_4$ $C{a}_3(P{O}_4{)}_2$ $K_4\left[Fe\right(CN{)}_6]$

Valency: $1$ $2$ $3$ $4$

The decreasing order is:

$\left(d\right)>\left(c\right)>\left(b\right)>\left(a\right)$