At what temperature would a 1.80 m nacl solution freeze, given that the van't hoff factor for nacl is 1.9? kf k_f for water is 1.86 degrees c/m. answer in degrees c

a 0.050 m solution of alcl3 had an observed osmotic pressure of 3.85 atm at 20 degrees c. calculate the van't hoff factor i ifor alcl3.alcl 3

Freezing point depression and osmotic pressure are colligative properties.

We know that;

ΔT = k m i

Where;

ΔT = freezing point depression

k = freezing point depression constant

m = molality of solution

i = Van't Hoff factor

Substituting values;

ΔT = 1.86 × 1.80 m × 1.9

ΔT = 6.36°C

Freezing point of pure solvent= 0°C

ΔT = Freezing point of pure solvent - Freezing point of solution

6.36°C =  0°C - Freezing point of solution

Freezing point of solution = 0°C - 6.36°C

= - 6.36°C

2) P = M R T i

P = osmotic pressure

M = molar concentration of solution

R = gas constant

T = temperature

i = Van't Hoff factor

i = P/ M R T

P = 3.85 atm

M = 0.050 M

R = 0.082 atm LK-1mol-1

T = 20°C + 273 = 293 K

i =  3.85 atm/0.050 M ×  0.082 atm LK-1mol-1  × 293 K

i = 3.2

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T°freezing solution = - 6.3°C

i = 3.20

Explanation:

This is all about colligatives properties.

First question:

- Freezing point depression ΔT = Kf . m . i

ΔT = T° freezing pure solvent - T° freezing solution

Kf = 1.86 °C/m

m = molality (mol of solute/kg of solvent)

i = Van't Hoff factor

(0° - (T°freezing solution) = 1.86 °C/m  . 1.80m . 1.9

T°freezing solution = - 6.3°C

Second question

Osmotic pressure = M . R . T. i

M is molarity

R is universal gases constant,  0.082 L.atm/mol.K

T is temperature in K (°C + 273) → 20°C + 273 = 293K

3.85 atm = 0.05mol/L . 0.082 L.atm/mol.K . 293K . i

3.85 atm / 0.05 . 0.082atm . 293 = i

3.20 = i

Theoretical i for AlCl₃ = 4

AlCl₃  →  1Al³⁺  +  3Cl⁻