SME
Grade 10 Science · Chemistry
Phase changes, vapor pressure and colligative properties
Use this for “27.75 g of CaCl₂ in 0.5 kg of water. What are the new freezing and boiling points?”
Diagram
Water’s phase diagram: pressure not to scale, the temperature axis broken, each part to its own scale.
At 1 atm pure water melts at 0 °C and boils at 100 °C.
The solution’s melting line moves left: it freezes at −2.79 °C, 2.79 °C lower.
Its boiling curve moves right: it boils at 100.768 °C, 0.768 °C higher.
Ice’s own line stays: the ice that forms is pure water.
This is an example. Type your own values to start a new problem.
g
Mass of solute
example · g
M
Molar mass
example · g/mol
n
Moles of solute
calculated · mol
w
Mass of water
example · kg
b
Molality
calculated · mol/kg
i
Particles each unit gives (van ’t Hoff factor)
example
ΔTf
Freezing point drop
calculated · °C
Tf
Freezing point of the solution
calculated · °C
ΔTb
Boiling point rise
calculated · °C
Tb
Boiling point of the solution
calculated · °C
Formulas
n
=
g
M
0.25 = 27.75/111
b
=
n
w
0.5 = 0.25/0.5
ΔTf
=
i
×
1.86
×
b
2.79 = 3 × 1.86 × 0.5
Tf = 0 − ΔTf
−2.79 = 0 − 2.79
ΔTb
=
i
×
0.512
×
b
0.768 = 3 × 0.512 × 0.5
Tb = 100 + ΔTb
100.77 = 100 + 0.768
Assumptions
•
The solvent is water: it freezes at 0 °C and boils at 100 °C, with Kf = 1.86 and Kb = 0.512 °C·kg/mol.
•
Only the number of dissolved particles matters: i is 1 for sugar, 2 for NaCl, 3 for CaCl₂, 4 for FeCl₃.
•
Molality b is moles of solute per kilogram of water, not per liter of solution.
•
The rule fits dilute solutions (up to about 1 mol/kg); stronger ones stray from it.
Step-by-step
WE KNOW
g = 27.75 g, M = 111 g/mol, w = 0.5 kg, i = 3
FIND
moles of solute (n), molality (b), freezing point drop (ΔTf), freezing point of the solution (Tf), boiling point rise (ΔTb), boiling point of the solution (Tb)
Step 1 · Find moles of solute (n)
Use n = g/M
Divide the grams by the grams in each mole.
n
=
27.75/111
n
=
0.25
mol
Step 2 · Find molality (b)
Use b = n/w
Molality is moles of solute per kilogram of water.
b
=
0.25/0.5
b
=
0.5
mol/kg
Step 3 · Find freezing point drop (ΔTf)
Use ΔTf = i × 1.86 × b
Water’s freezing point changes 1.86 °C for each mole of particles per kilogram.
ΔTf = 3 × 1.86 × 0.5
ΔTf = 5.58 × 0.5
ΔTf = 2.79 °C
Step 4 · Find freezing point of the solution (Tf)
Use Tf = 0 − ΔTf
Pure water freezes at 0 °C; the solute pushes it lower.
Tf = 0 − 2.79
Tf = −2.79 °C
Step 5 · Find boiling point rise (ΔTb)
Use ΔTb = i × 0.512 × b
Water’s boiling point changes 0.512 °C for each mole of particles per kilogram.
ΔTb = 3 × 0.512 × 0.5
ΔTb = 1.536 × 0.5
ΔTb = 0.768 °C
Step 6 · Find boiling point of the solution (Tb)
Use Tb = 100 + ΔTb
Pure water boils at 100 °C; the solute pushes it higher.
Tb = 100 + 0.768
Tb = 100.77 °C
Check
Put the numbers back in. Both sides should match.
0.25 = 27.75/111 ✓
0.5 = 0.25/0.5 ✓
2.79 = 3 × 1.86 × 0.5 ✓
−2.79 = 0 − 2.79 ✓
0.768 = 3 × 0.512 × 0.5 ✓
100.77 ≈ 100 + 0.768 ✓
Topic check
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