Question

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at −14.5 ∘C

Answer 1

Answer:

108.43 grams KNO₃

Explanation:

To solve this problem we use the formula:

• ΔT = Kf * b * i

Where

• ΔT is the temperature difference (14.5 K)

• Kf is the cryoscopic constant (1.86 K·m⁻¹)

• b is the molality of the solution (moles KNO₃ per kg of water)

• and i is the van't Hoff factor (2 for KNO₃)

We solve for b:

• 14.5 K = 1.86 K·m⁻¹ * b * 2

• b = 3.90 m

Using the given volume of water and its density (aprx. 1 g/mL) we calculate the necessary moles of KNO₃:

• 275 mL water ≅ 275 g water

• 275 g /1000 = 0.275 kg

• moles KNO₃ = molality * kg water = 3.90 * 0.275

• moles KNO₃ = 1.0725 moles KNO₃

Finally we convert KNO₃ moles to grams, using its molecular weight:

• 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃