Question

Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?

पर्याय

• NaCl

• Na₂SO₄

• C₆H₁₂O₆

• Al₂(SO₄)₃

Answer 1

Al₂(SO₄)₃

Explanation:

Freezing point depression is given by the formula;

ΔT_f = i × K_b × m

Where, ΔT_f = Freezing point depression

T_f = Freezing point of the solution (°C)

i = ionization or van't Hoff factor

K_f = 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.