At –10.5 °C (a common temperature for household freezers), what is the maximum m

Question

At –10.5 °C (a common temperature for household freezers), what is the maximum mass of fructose (C6H12O6) you can add to 1.00 kg of pure water and still have the solution freeze? Assume that fructose is a molecular solid and does not ionize when it dissolves in water. Kf values are given here.

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)

Solvent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

Kb value

(°C/m)

Normal boiling

point (°C)

water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon
tetrachloride   CCl4 29.8 –22.9 5.03 76.8 camphor   C10H16O 37.8 176

Explanation / Answer

depression in freezing point = freezing point of pure water - freezing point of solution

depression in freezing point = 0 - (-10.5)

depression in freezing point = 10.5 C

now

we know that

depression in freezing point = Kf x molality

Kf for water = 1.86 C/ m

so

10.5 = 1.86 x molality

molality = 5.645 m

now

molality = moles of solute / mass of water (kg)

5.645 = moles of solute / 1

moles of solute = 5.645 mol

now

mass of solute fructose = moles x molar mass

mass of solute fructose = 5.645 mol x 180 g/mol

mass of solute fructose = 1016.13 g

so

1016.13 grams of fructose can be added

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.