In an electrical field, the particles of a colloid system move towards cathode. The coagulation of the same sol is studied using K2SO4(I), Na3PO4 (II), K4[Fe(CN)6] (III) and NaCl (IV). The coagulating power should be

Question

In an electrical field, the particles of a colloid system move towards cathode. The coagulation of the same sol is studied using K2SO4(I), Na3PO4 (II), K4[Fe(CN)6] (III) and NaCl (IV). The coagulating power should be (A) I > II > III > IV (B) III > II > I > IV (C) III > I > II > IV (D) IV > III > I > II

Tardigrade Admin · Accepted Answer

The given sol is positively charged because the colloidal particles move towards cathode. Its coagulation can be caused by oppositely charged anion therefore, the coagulating power is

[F e (CN)_{6}]^{4 -} > P O_{4}^{3 -} > S O_{4}^{2 -} > C l^{-}

Q. In an electrical field, the particles of a colloid system move towards cathode. The coagulation of the same sol is studied using K2​SO4​(I),Na3​PO4​(II),K4​[Fe(CN)6​](III) and NaCl(IV). The coagulating power should be

I > II > III > IV

III > II > I > IV

III > I > II > IV

IV > III > I > II

The given sol is positively charged because the colloidal particles move towards cathode. Its coagulation can be caused by oppositely charged anion therefore, the coagulating power is [Fe(CN)6​]4−>PO43−​>SO42−​>Cl−

Q. In an electrical field, the particles of a colloid system move towards cathode. The coagulation of the same sol is studied using $K_{2} S O_{4} (I), N a_{3} P O_{4} (II), K_{4} [F e (CN)_{6}] (III)$ and $N a Cl (I V)$ . The coagulating power should be

The given sol is positively charged because the colloidal particles move towards cathode. Its coagulation can be caused by oppositely charged anion therefore, the coagulating power is
$[F e (CN)_{6}]^{4 -} > P O_{4}^{3 -} > S O_{4}^{2 -} > C l^{-}$