Question

What mass of salt (NaCl) should you add to 1.04 L of water in an ice cream maker to make a solution that freezes at -10.8 ∘C ? Assume complete dissociation of the NaCl and density of 1.00 g/mL for water.

Answer 1

176.5 g is the mass of needed NaCl

Step-by-step explanation:

Colligative property of freezing point depression is needed to be applied here:

Freezing T° pure solvent - Freezing T° of solution = Kf . m . i

i = 2, since we assume, the NaCl complete dissociation

NaCl → Na⁺ + Cl⁻

If the density of water is 1.04 g/mL, we convert the volume of water from L to mL in order to get water's mass

1.04L . 1000mL / 1L = 1040 mL

Water mass = Water density . water volume → 1 g/mL . 1040 mL = 1040 g

In order to work with molality, we convert the mass from g to kg

1040 g. 1kg / 1000 g = 1.04 kg.

Let's replace data → 0°C - (-10.8°C ) = 1.86 °C/m . m . 2

10.8°C/ 1.86 m/°C . 2 = m → 2.90 m

This represents molality, moles of solute in 1kg of solvent. But our mass of solvent is 1.04 kg, so let's find the moles that corresponds to that:

Molality . mass of solvent → 2.90 mol/kg . 1.04kg = 3.02 moles

To reach the answer, we convert the moles to mass:

3.02 mol . 58.45 g / 1mol = 176.5 g