Question:medium

A colloidal solution is subjected to an electric field. The colloidal particles move toward the anode. In the coagulation of this solution using NaCl, BaCl$_2$ and AlCl$_3$ separately, which one is the correct order of coagulation power?

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Remember the Hardy-Schulze rule: for coagulation of colloids, the effective ions are those with charge opposite to the colloidal particles. Higher the valency of these effective ions, greater their coagulating power.
Updated On: May 6, 2026
  • NaCl \textgreater BaCl$_2$ \textgreater AlCl$_3$
  • BaCl$_2$ \textgreater AlCl$_3$ \textgreater NaCl
  • NaCl \textgreater AlCl$_3$ \textgreater BaCl$_2$
  • AlCl$_3$ \textgreater BaCl$_2$ \textgreater NaCl
Show Solution

The Correct Option is D

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