Question

Which of the following aqueous solutions has the lowest freezing point? The Kf of water is 1.86 kg.oC/mol.

a. 0.8 m CH3OH (methanol)

b. 0.4 m NaCl

c. 0.3 m MgBr2

d. 0.2 m AlF3

e. 0.25 m Ca(NO3)2

Calculate the freezing point ONLY for the solution you marked as your answer above.

Answer 1

Ans: C

Solution:

Colligative property is dependent on no of solute particles i.e. no of ions/species.

Species with highest value of (molality x no of ions/species) has highest Depression in freezing point which in turn has lowest freezing point. ( Depression in freezing point is a colligative property )

a) 0.8 m CH₃OH

= 0.8 x 1 ( CH₃OH is a non-electrolyte)

= 0.8

b) 0.4 m NaCl

= 0.4 x 2 ( NaCl = Na⁺ + Cl^- = 2 ions)

= 0.8

c) 0.3 m MgBr₂

= 0.3 x 3 ( MgBr₂ = Mg²⁺ + 2Br^- = 3 ions)

= 0.9

d) 0.2 m AlF₃

= 0.2 x 4 (AlF₃ = Al³⁺ + 3F^- = 4 ions)

= 0.8

e) 0.25 m Ca(NO₃)₂

= 0.25 x 3 [Ca(NO₃)₂ = Ca⁺² + 2NO₃^- = 3 ions]

= 0.75

Therefore, 0.3 m MgBr₂ has highest Depression in freezing point which in turn has lowest freezing point.

Hence, 0.3 m MgBr₂ has the lowest freezing point.