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  • Among the electrolytes <img alt="\mathrm{Na}_2 \mathrm{SO}_4, \mathrm{CaCl}_2, \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ and $\mathrm{NH}_4 \mathrm{Cl}" src="https://entrancecorner.oncodecogs.com/

Among the electrolytes \mathrm{Na}_2 \mathrm{SO}_4, \mathrm{CaCl}_2, \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ and $\mathrm{NH}_4 \mathrm{Cl} the most effective coagulating agent for \mathrm{\mathrm{Sb}_2 \mathrm{S}_3} sol is:
 

Option: 1

\mathrm{Na}_2 \mathrm{SO}_4


Option: 2

\mathrm{CaCl}_2


Option: 3

\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3
 


Option: 4

\mathrm{NH}_4 \mathrm{Cl}


Answers (1)

best_answer

Hardy-Schulze Rule: Greater the valence of flocculating ion added, greater is its power to cause coagulation. \mathrm{\mathrm{Sb}_2 \mathrm{S}_3} is negative sol. So, cation with greater valency will have greater power to cause coagulation. Among the given electrolytes, \mathrm{\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3} have the higher valence cation \mathrm{\left(\mathrm{Al}^{3+}\right)}. Therefore, \mathrm{\mathrm{Al}^{3+}} has highest coagulation power for \mathrm{\mathrm{Sb}_2 \mathrm{S}_3}
Order of coagulating power of given cations is
\mathrm{ \mathrm{Al}^{3+}>\mathrm{Ca}^{2+}>\mathrm{Na}^{+} \equiv \mathrm{NH}_4^{+} }

Hence option 3 is correct.

Posted by

Nehul

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