Question

Which one of the following is most effective in causing the coagulation of an $A{s}_2{S}_3$ sol ?

• A. $KCl$
• B. $AlC{l}_3$
• C. $MgS{O}_4$
• D. $K_{3}Fe(CN{)}_6$

Answer 1

The correct option is B $AlC{l}_3$

According to Hardy Schulze rule, greater the valency of the coagulating ion, greater is its coagulating power.

$A{s}_2{S}_3$ is a negative sol. Thus, out of the given compound, $AlC{l}_3\left(A{l}^{3+}\right)$ is most effective for causing a coagulation of $A{s}_2{S}_3$ sol.

Hence, option $B$ is correct.