Answer to Question #299838 in Chemistry for Wellen

Question

How much table salt should be added to 300 g water to lower its freezing point to -2.00°C? 

Answer 1

Solution:

table salt = sodium chloride = NaCl

NaCl → Na⁺ + Cl⁻

Sodium chloride (NaCl) consists of two ions i.e. the sodium ion (Na⁺) and the chlorine ion (Cl⁻).

So, the Van't Hoff factor for NaCl considering complete dissociation (α = 100%) is 2.

The lowering (depression) of the freezing point of the solvent can be represented using the following equation: 

Δt = i × K_f × m

where:

Δt = the change in freezing point

i = Vant Hoff factor

K_f = the freezing point depression constant

m = the molality of the solute

The freezing point of pure water is 0°C.

Hence, Δt = 0°C − (−2.00°C) = 2.00°C

i = 2 (for NaCl)

K_f = 1.86°C/m (for water)

Therefore,

m = Δt / (i × K_f)

Molality of NaCl = (2.00°C) / (2 × 1.86°C/m) = 0.5376 m

Molality = Moles of solute / Kilograms of solvent

Therefore,

Moles of NaCl = Molality of NaCl × Kilograms of water

Moles of NaCl = 0.5376 mol kg⁻¹ × 0.3 kg = 0.1613 mol

The molar mass of NaCl is 58.44 g/mol

Therefore,

Mass of NaCl = (0.1613 mol NaCl) × (58.44 g NaCl / 1 mol NaCl) = 9.426 g NaCl = 9.43 g NaCl

Mass of NaCl = 9.43 g

Answer: 9.43 grams of table salt (NaCl) should be added.