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

Respuesta :

Answer:

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

Mgā‚ƒ(POā‚„)ā‚‚ - Ā AlBrā‚ƒ - BeBrā‚‚ - Ā KBr

Explanation:

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

Formula is:

ΔT = Kf . m . i

Where Ī”T = 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:

BeBrā‚‚  → Be²⁺ Ā + Ā 2Br⁻ Ā  i = 3

AlBrā‚ƒ → Ā Al³⁺ Ā + 3Br⁻ Ā  i = 4

Mgā‚ƒ(POā‚„)ā‚‚  → Ā 3Mg²⁺ Ā + Ā 2PO₄⁻³ Ā  Ā i = 5

KBr → Ā K⁺ Ā + Ā Br⁻ Ā  i = 2

Solution of Mgā‚ƒ(POā‚„)ā‚‚ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).