Question

At what temperature would a 1.30 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 ∘C/m .

Answer 1

Answer:

-4.59°C

Explanation:

Let's see the formula for freezing point depression.

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Freezing constant. For water if 1.86°C/m

m = molality (moles of solute in 1kg of solvent)

i = Van't Hoff factor.

0°C - Freezing T° of solution = 1.86°C /m . 1.30m . 1.9

Freezing T° of solution = - (1.86°C /m . 1.30m . 1.9)

Freezing T° of solution = - (1.86°C/m .  1.30m . 1.9) → -4.59°C

NaCl →  Na⁺  +  Cl⁻