Question

Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freezing point to highest freezing point.
BeBr2, AlBr3, Mg3(PO4)2, KBr

Answer 1

Answer:

The solutions are listed from the lowest freezing point to highest freezing point.

Mg₃(PO₄)₂ -  AlBr₃ - BeBr₂ -  KBr

Explanation:

Colligative property of freezing point, is the freezing point depression.

Formula is:

ΔT = Kf . m . i

Where ΔT = T° freezing pure solvent - T° freezing solution

Kf is the cryoscopic constant

m is molality and i is the Van't Hoff factor (number of ions, dissolved in solution)

In this case, T° freezing pure solvent is 0° because it's water, so the Kf is 1.86 °C/m, and m is 0.1 molal.

The i modifies the T° freezing solution. The four salts, are ionic compounds, so for each case:

BeBr₂  → Be²⁺  +  2Br⁻   i = 3

AlBr₃ →  Al³⁺  + 3Br⁻   i = 4

Mg₃(PO₄)₂  →  3Mg²⁺  +  2PO₄⁻³    i = 5

KBr →  K⁺  +  Br⁻   i = 2

Solution of Mg₃(PO₄)₂ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).