Question

Which of the following solutions will freeze at the lowesttemperature? a.1 mole C6H12O6in 500 g water b.1 mole MgF2in 500 g water c.1 mole KBr in 500 g water d.1 mole AlCl3in 500 g water

Answer 1

Answer:

The compoud AlCl3 will form 4 particles, so this solution will be the lowest freezing point temperature. Option D is correct.

Explanation:

Step 1: Data given

mass of water = 500 grams = 0.500 kg

We have 1 mol of every solute

Step 2: Calculate the freezing point temperature

ΔT = i*Kf * m

⇒ΔT = the freezing point depression = The solution with the biggest freezing point depression, will have the lowest freezing temperature

⇒i = the van't Hoff factor = Says, after solving in water, in how many particles the compound will form

⇒Kf = the freezing point depression constant of water = 1.86 °C/m

⇒m = the molalty = 1 mol / 0.500 kg = 2 molal

a.1 mole C6H12O6in 500 g water

ΔT = 1*1.86*2

ΔT = 3.72 °C

The freezing point is 0 -3.72 °C = -3.72 °C

b.1 mole MgF2in 500 g water

ΔT = 3*1.86*2

ΔT = 11.16 °C

The freezing point is 0 -11.16 °C = -11.16 °C

c.1 mole KBr in 500 g water

ΔT = 2*1.86*2

ΔT = 7.44 °C

The freezing point is 0 -7.44 °C = -7.44 °C

d.1 mole AlCl3in 500 g water

ΔT = 4*1.86*2

ΔT = 14.88 °C

The freezing point is 0 -14.88 °C = -14.88 °C

The compoud AlCl3 will form 4 particles, so this solution will be the lowest freezing point temperature. Option D is correct.

Answer 2

Answer:

Option d.

1 mole AlCl3in 500 g water

Explanation:

ΔT = Kf . m . i

Freezing T° of solution = - (Kf . m . i)

In order to have the lowest freezing T° of solution, we need to know which solution has the highest value for the product (Kf . m . i)

Kf is a constant, so stays the same and m stays also the same because we have the same moles, in the same amount of solvent. In conclussion, same molality to all.

i defines everything. The i refers to the Van't Hoff factor which are the number of ions dissolved in solution. We assume 100 & of ionization so:

a. Glucose → i = 1

Glucose is non electrolytic, no ions formed

b. MgF₂ →  Mg²⁺ + 2F⁻

i = 3. 1 mol of magnessium cation and 2 fluorides.

c. KBr  →  K⁺ + Br⁻

i = 2. 1 mol potassium cation and 1 mol of bromide anion

d. AlCl₃ →  Al³⁺ + 3Cl⁻

i = 4. 1 mol of aluminum cation and 3 mol of chlorides.

Kf . m . 4 → option d will has the highest product, therefore will be the lowest freezing point.