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.
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.
