Among the electrolytes Na_2SO_4,CaCl_2,Al_2(S | vedclass

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(C) According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the active ion (the ion carrying a charge op

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$Al_2(SO_4)_3$

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(C) According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the active ion (the ion carrying a charge opposite to that of the colloidal particles).$Sb_2S_3$ sol is a negatively charged sol. Therefore,it is coagulated by the positively charged ions of the added electrolytes.The positively charged ions in the given electrolytes are:$Na^+$ (from $Na_2SO_4$),$Ca^{2+}$ (from $CaCl_2$),$Al^{3+}$ (from $Al_2(SO_4)_3$),$NH_4^+$ (from $NH_4Cl$).According to the Hardy-Schulze rule,the coagulating power increases with the increase in the magnitude of the charge on the ion. The order of coagulating power is $Al^{3+} > Ca^{2+} > Na^+ = NH_4^+$.Since $Al^{3+}$ has the highest charge $(+3)$,$Al_2(SO_4)_3$ is the most effective coagulating agent for $Sb_2S_3$ sol.Hence,option $C$ is correct.

Among the electrolytes $Na_2SO_4$,$CaCl_2$,$Al_2(SO_4)_3$,and $NH_4Cl$,the most effective coagulating agent for $Sb_2S_3$ sol is

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