SME
General Chemistry II
Solutions and colligative properties
Use this for “3.00 g of a molecular solid in 50 g of water lowers the freezing point by 0.620 °C. Find its molar mass.”
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 −0.62 °C, 0.62 °C lower.
Its boiling curve moves right: it boils at 100.1707 °C, 0.1707 °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.
m
Solute mass
example · g
mw
Water mass
example · g
i
Van ’t Hoff factor
example
Kf
Freezing-point constant
example · °C·kg/mol
ΔTf
Freezing-point depression
example · °C
b
Molality
calculated · mol/kg
n
Moles of solute
calculated · mol
M
Molar mass
calculated · g/mol
Tf
Freezing point
worked out (not typed) · °C
Tb
Boiling point in water
worked out (not typed) · °C
Formulas
ΔTf
=
i
×
Kf
×
b
0.62 = 1 × 1.86 × 0.3333
b
=
1,000
×
n
mw
0.3333
=
1,000
×
0.01667
50
M
=
m
n
180
=
3
0.01667
Tf = 0 − ΔTf
−0.62 = 0 − 0.62
Tb
=
100
+
0.512
×
i
×
b
100.2 = 100 + 0.512 × 1 × 0.3333
Assumptions
•
The solution is dilute and the solute does not evaporate, so ΔTf = i × Kf × b, with b the moles of solute per kilogram of water.
•
The van ’t Hoff factor i counts the particles from each formula unit: 1 for a molecule, 2 for NaCl, taken as ideal.
•
Kf belongs to the solvent: water’s is 1.86 °C·kg/mol. The picture draws water’s phase diagram, with the boiling point the same solute gives water (Kb = 0.512 °C·kg/mol).
Step-by-step
WE KNOW
m = 3 g, mw = 50 g, ΔTf = 0.62 °C, i = 1, Kf = 1.86 °C·kg/mol
FIND
molality (b), moles of solute (n), molar mass (M), freezing point (Tf), boiling point in water (Tb)
Step 1 · Find molality (b)
Use ΔTf = i × Kf × b
Divide
the
drop
by
i
×
Kf:
the
molality
of
particles,
shared
out
per
formula
unit.
b
=
ΔTf
i
× Kf
b
=
0.62
1 × 1.86
b
=
0.3333
mol/kg
Step 2 · Find moles of solute (n)
Use b = 1,000 × n ÷ mw
Molality
times
the
water
in
kilograms
(
g
1000
)
counts
the
moles
of
solute.
n
=
b
×
mw
1,000
n
=
0.3333
×
50
1,000
n
=
16.665
1,000
n
=
0.01667
mol
Step 3 · Find molar mass (M)
Use M = m ÷ n
Grams of solute over its moles is the molar mass.
M
=
3
0.01667
M
=
180
g/mol
Step 4 · Find freezing point (Tf)
Use Tf = 0 − ΔTf
Pure water freezes at 0 °C; the solute pushes it lower.
Tf = 0 − 0.62
Tf = −0.62 °C
Step 5 · Find boiling point in water (Tb)
Use Tb = 100 + 0.512 × i × b
Pure water boils at 100 °C; the solute pushes it higher, by Kb = 0.512 °C·kg/mol for each mole of particles per kilogram.
Tb = 100 + 0.512 × 1 × 0.3333
Tb = 100 + 0.1706
Tb = 100.2 °C
Check
Put the numbers back in. Both sides should match.
0.62 = 1 × 1.86 × 0.3333 ✓
0.3333
=
1,000
×
0.016667
50
✓
180
=
3
0.01667
✓
−0.62 = 0 − 0.62 ✓
100.2 = 100 + 0.512 × 1 × 0.3333 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.