Answer:
-3.82ºC is the freezing point of solution
Step-by-step explanation:
We work with the Freezing point depression to solve the problem
ΔT = m . Kf . i
ΔT = Freezing point of pure solvent - freezing point of solution
Let's find out m, molality (moles of solute in 1kg of solvent)
15 g / 58.45 g/mol = 0.257 moles of NaCl
NaCl(s) → Na⁺ (aq) + Cl⁻(aq)
i = 2 (Van't Hoff factor, numbers of ions dissolved)
m = mol /kg → 0.257 mol / 0.250kg = 1.03 m
Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)
0ºC - Tºf = 1.03m . 1.86ºC/m . 2
Tºf = -3.82ºC