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

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°).

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