Of the following $$0.10$$ m aqueous solutions, which one will exhibit the largest freezing point depression?

Question

Infinity Learn · Accepted Answer

Freezing point depression $$($$∆$$Tf)$$ is a colligative property and depends upon the van't Hoff factor (i), i.e. number of ions given by the electrolyte in aqueous solution.               ∆$$Tf=i$$×Kf×mWhere, $$Kf=molal$$ freezing point depression constant              $$m=$$ molality of the solution∴Kf and m are constant ∴∆Tf∝$$i(a)$$ $$KCl(aq)$$⇌$$K+(aq)+Cl-(aq)(total$$ ions $$=2$$ thus,$$i=2)(b)$$ $$C2H12O6$$⇌no ions [$$i=0$$](c$$ $$) $$Al2SO43(aq)$$⇌$$2Al3++3SO42-$$                                        (total$$ ions $$-5$$, thus, i $$=$$ $$5)(d)K2SO4(aq)⇌$$2K++SO42-(total$$ ions $$-$$ $$3$$,thus, i $$=$$ $$3)Hence$$, $$Al2(SO4)3$$ will exhibit largest freezing point depression due to the highest value of i.

Q.

Unlock the Full Solution and Master the Concept