Which of the following is most powerful to coagulate the negative colloid?
A.
$$ZnS{O_4}$$
B.
$$N{a_3}P{O_4}$$
C.
$$AlC{l_3}$$
D.
$${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$
Answer :
$$AlC{l_3}$$
Solution :
According to Hardy-Schulze rule "The amount of electrolyte required to coagulate a fixed amount of a sol depends upon the sign of charge and valency of the flocculating ion."
Thus, the coagulating power vary in the order.
$$A{l^{3 + }} > Z{n^{2 + }} > N{a^ + }$$
Releted MCQ Question on Physical Chemistry >> Surface Chemistry