Urgent! if a solution is made by dissolving 3.00 grams of magnesium chlorite into 30.0 grams of acetic acid (c2h4o2), what's the freezing point? (kf of acetic acid = 3.90 °c/m; freezing point of pure acetic acid = 16.6 °c)

Change in freezing point = Kf * (molality)* i

1) Kf for acetic acid = 3.90 C/molal

2) Molality = moles solute/kilograms

3.00 grams Mg(ClO2)2/ 30.0 g acetic acid x 1000 g/kg x 1mole Mg(ClO2)2 /159.2 grams = 0.628

3) i is the van’t hoff factor ;

Mg(ClO2)2 ; 24.3 + 2(35.45) + 4(16) = 159.2

Mg(ClO2)2 ==⇒ Mg2+ + 2ClO2-

The value of i is 3, since 3 ions are produced

Change in freezing point = 3.90 C/molal x 0.628 x 3 = 7.35 Celsius

Freezing point of solution = 16.6 – 7.35 = 9.3 Celsius

it is stable as it is in group 0 which have full outer shells of electrons

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