Among the electrolytes ${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}{\rm{,CaC}}{{\rm{l}}_2}{\rm{,A}}{{\rm{l}}_2}{\left( {{\rm{S}}{{\rm{O}}_4}} \right)_3}\;{\rm{and}}\;{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$, the most effective coagulation agent for ${\rm{S}}{{\rm{b}}_2}{{\rm{S}}_3}$ sol is

${\rm{A}}{{\rm{l}}_2}{\left( {{\rm{S}}{{\rm{O}}_4}} \right)_3}$

${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$

${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$

NEET UG CHEMISTRY MCQ: SURFACE CHEMISTRY

1 Easy

1 of 1

Application +4 | -1

Among the electrolytes

\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}\mathrm{,CaC}{\mathrm{l}_2}\mathrm{,A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}\;\mathrm{and}\;\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

, the most effective coagulation agent for

\mathrm{S}{\mathrm{b}_2}{\mathrm{S}_3}

sol is

View answer and explanation

Correct option: C

\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}

Option C (

\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}

) is correct because, according to the Hardy-Schulze rule, the ion with the highest charge will be the most effective in coagulating a sol. Since

Sb_2S_3

is a negative sol, the cation with the highest charge (

Al^{3+}

) will be the most effective.