# A chemistry question.?

If a solution is made by dissolving 3.00 grams of magnesium chlorite into 30.0 grams of acetic acid (C2H4O2), what is the freezing point? (Kf of acetic acid = 3.90 °C/m; freezing point of pure acetic acid = 16.6 °C)

Relevance
• To solve this problem, we need to determine the number of moles of magnesium chlorite.

According to the website above, the formula of this compound is Mg(ClO2)2. The mass of one mole is 159.2086 grams. To determine the number of moles of this compound, divide 3 grams by this number.

n = 3 ÷ 159.2086

This is approximately 0.0063 mole. Each mole contains three ions.

n = 9 ÷ 159.2086

The next step is to determine the number of moles of the ions by the number of kilograms of the solvent.

m = (9 ÷ 159.2086) ÷ 0.03 = 300 ÷ 159.2086

This is approximately 1.88. To determine the decrease of the freezing point, multiply this number by the Kf.

∆ T = 3.90 * 300 ÷ 159.2086 = 1170÷ 159.2086

This is approximately 7.35˚.

Tf = 16.6 – (1170÷ 159.2086)

This is approximately 9.25˚C.

• Change in freezing point = Kf(molality)i

Kf for acetic acid = 3.90 C/molal

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

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

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