What mass of salt (nacl) should you add to 1.48 l of water in an ice cream maker to make a solution that freezes at -13.4 ∘c ? assume complete dissociation of the nacl and density of 1.00 g/ml for water?

Answer is: mass of salt is 311,15 g.
V(H₂O) = 1,48 l · 1000 ml/l = 1480 ml.
m(H₂O) = 1480 g = 1,48 kg.
d(solution) = 1,00 g/ml.
ΔT(solution) = 13,4°C = 13,4 K.
Kf = 1,86 K·kg/mol; cryoscopic constant of water
i(NaCl) = 2; Van 't Hoff factor.
ΔT(solution) = Kf · b · i.
b(NaCl) = 13,4 K ÷ (1,86 K·kg/mol · 2).
b(NaCl) = 3,6 mol/kg.
n(NaCl) = 3,6 mol · 1,48 kg= 5,328 mol.
m(NaCl) = 5,328 mol · 58,4 g/mol = 311,15 g.

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jufsanabriasa jufsanabriasa · Answer 2 · 2020-04-20T03:34:58+02:00

Answer:

311.6g NaCl you should add

Explanation:

When you add a solute (NaCl) to solvent (Water), the freezing point of the solution decreases with regard to pure solvent following the equation:

ΔT = Kf × m × i

Where ΔT is change in temperature(From 0°C to -13.4°C), Kf is freezing point depression constant (1.86°C/m for water), m is molality of solution (Moles solute / 1.48 kg solvent -Because 1.48L≡1.48kg; density 1.00g/mL-) and i is Van't Hoff factor (2 for NaCl because in water, NaCl dissociates as Na⁺ and Cl⁻ ions, 2 ions).

Replacing:

13.4°C = 1.86°C/m × moles NaCl / 1.48kg × 2

5.33 = moles NaCl

As molar mass of NaCl is 58.44g/mol, mass in 5.33moles are:

5.33mol NaCl × (58.44g /mol) = 311.6g NaCl you should add