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Question
Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?
Options
NaCl
Na2SO4
C6H12O6
Al2(SO4)3
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Solution
Al2(SO4)3
Explanation:
Freezing point depression is given by the formula;
ΔTf = i × Kb × m
Where, ΔTf = Freezing point depression
Tf = Freezing point of the solution (°C)
i = ionization or van't Hoff factor
Kf = cryoscopic constant (°C/m)
m = molality (mol/kg)
Let's check for each option one by one.
(1) \[\ce{KCl_{(aq)} <=> K^+_{ (aq)} + Cl^-_{ (aq)}}\]
total ions = 2 thus, i = 2
(2) Glucose is non-electrolyte, so it does not dissociate into ions.
\[\ce{C6H12O6 <=> no ions}\] [i = 0]
(3) \[\ce{Al2(SO4)3_{(aq)} <=> 2Al^{3+} + 3SO^{2-}_4}\]
total ions = 5 thus, i = 5
(4) \[\ce{K2SO4_{ (aq)} <=> 2K^+ + SO^{2-}_4}\]
total ions = 3 thus, i = 3
Hence, Al2(SO4)3 will exhibit the largest freezing point depression due to the highest value of i.
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