What mass of salt (NaCl) should you add to 1.62 L of water in an ice-cream maker to make a solution that freezes at -11.8 ∘C? Assume complete dissociation of the NaCl and a density of 1.00 g/mL for water and use Kf=1.86∘C/m.

300.4 g of salt must be added.

Explanation:

Let's replace the data given in the formula for the colligative property of freezing point depression.

0°C - (-11.8°C) = 1.86°C /m . m . 2

i = 2, because we assumed complete dissociation of the solute, NaCl

NaCl →  Na⁺  + Cl⁻    i = 2

i = Van't Hoff factor, ions dissolved in solution

Let's find out the m (molality) . Molality are the moles of solute in 1kg of solvent.

We need the mass of solvent, so we work with density.

With molality and the mass of solvent, we can determine the moles of solute. Therefore, we can know its mass

1g/mL = Solvent mass / Solvent volume

1 g/mL . 1620 mL = 1620 g

(Be careful, we converted the volume from L to mL, → 1.62L . 1000mL /L

11.8°C / 1.86 m/°C . 2 = m → 3.17 mol/kg

In 1kg of solvent we have 3.17 moles of solute

Our mass of solvent is 1620 g → 1620 g . 1kg/1000 g = 1.62kg

Molality = Mol/kg → Molality . kg = Mol

3.17 mol/kg . 1.62 kg = 5.14 moles

Finally, we convert the moles to mass → 5.14 mol . 58.45 g /1mol = 300.4 g