In each of four separate beakers (I, II, III, | vedclass

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(A) $Fe_2O_3 \cdot xH_2O$ is a positively charged sol. According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the vale

Vedclass · Accepted Answer

$II > III > IV > I$

Vedclass · Answer

(A) $Fe_2O_3 \cdot xH_2O$ is a positively charged sol. According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the oppositely charged ion (anion in this case). The higher the valency of the flocculating ion,the greater is its coagulating power. The anions provided are: $Cl^-$ (valency $1$),$[Fe(CN)_6]^{4-}$ (valency $4$),$PO_4^{3-}$ (valency $3$),and $SO_4^{2-}$ (valency $2$). Comparing the valencies: $4 > 3 > 2 > 1$. Thus,the order of coagulating power is: $[Fe(CN)_6]^{4-} > PO_4^{3-} > SO_4^{2-} > Cl^-$. This corresponds to the beakers: $II > III > IV > I$.

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