Boiling point elevation and freezing point depression: Class 12 practice
ΔT_b = iK_b m and ΔT_f = iK_f m, where i is the van 't Hoff factor (1 for non-electrolytes, about 2 for NaCl, 3 for CaCl₂). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Molality
The idea behind boiling point elevation and freezing point depression
ΔT_b = iK_b m and ΔT_f = iK_f m, where i is the van 't Hoff factor (1 for non-electrolytes, about 2 for NaCl, 3 for CaCl₂). Water: K_b = 0.52 K kg/mol, K_f = 1.86 K kg/mol.
Step-by-step method
- Find molality m.
- Choose i.
- ΔT = iKm; new boiling point = 373.15 + ΔT_b, freezing point = 273.15 − ΔT_f.
Worked example 1 (Basic)
Find the freezing and boiling points of a 0.2 m aqueous glucose solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- i = 1
- ΔT_f = 1 × 1.86 × 0.2 = 0.372 K
- ΔT_b = 1 × 0.52 × 0.2 = 0.104 K
- Answer: f.p. -0.372 °C, b.p. 100.104 °C
Worked example 2 (Standard)
Find the freezing and boiling points of a 0.1 m aqueous NaCl solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- i = 2
- ΔT_f = 2 × 1.86 × 0.1 = 0.372 K
- ΔT_b = 2 × 0.52 × 0.1 = 0.104 K
- Answer: f.p. -0.372 °C, b.p. 100.104 °C
Worked example 3 (Challenge)
Find the freezing and boiling points of a 0.9 m aqueous CaCl₂ solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- i = 3
- ΔT_f = 3 × 1.86 × 0.9 = 5.022 K
- ΔT_b = 3 × 0.52 × 0.9 = 1.404 K
- Answer: f.p. -5.022 °C, b.p. 101.404 °C
Common mistakes and misconceptions
- Forgetting i for electrolytes
- Adding ΔT_f instead of subtracting
Practice questions
- (Basic) Find the freezing and boiling points of a 0.9 m aqueous glucose solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- (Basic) Find the freezing and boiling points of a 0.5 m aqueous glucose solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- (Standard) Find the freezing and boiling points of a 0.4 m aqueous NaCl solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- (Standard) Find the freezing and boiling points of a 1 m aqueous NaCl solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- (Challenge) Find the freezing and boiling points of a 0.5 m aqueous CaCl₂ solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
- (Challenge) Find the freezing and boiling points of a 1 m aqueous CaCl₂ solution (K_f = 1.86, K_b = 0.52 K kg/mol, complete dissociation).
Answer key with full working
- Q1: i = 1 → ΔT_f = 1 × 1.86 × 0.9 = 1.674 K → ΔT_b = 1 × 0.52 × 0.9 = 0.468 K → Answer: f.p. -1.674 °C, b.p. 100.468 °C
- Q2: i = 1 → ΔT_f = 1 × 1.86 × 0.5 = 0.93 K → ΔT_b = 1 × 0.52 × 0.5 = 0.26 K → Answer: f.p. -0.93 °C, b.p. 100.26 °C
- Q3: i = 2 → ΔT_f = 2 × 1.86 × 0.4 = 1.488 K → ΔT_b = 2 × 0.52 × 0.4 = 0.416 K → Answer: f.p. -1.488 °C, b.p. 100.416 °C
- Q4: i = 2 → ΔT_f = 2 × 1.86 × 1 = 3.72 K → ΔT_b = 2 × 0.52 × 1 = 1.04 K → Answer: f.p. -3.72 °C, b.p. 101.04 °C
- Q5: i = 3 → ΔT_f = 3 × 1.86 × 0.5 = 2.79 K → ΔT_b = 3 × 0.52 × 0.5 = 0.78 K → Answer: f.p. -2.79 °C, b.p. 100.78 °C
- Q6: i = 3 → ΔT_f = 3 × 1.86 × 1 = 5.58 K → ΔT_b = 3 × 0.52 × 1 = 1.56 K → Answer: f.p. -5.58 °C, b.p. 101.56 °C
Syllabus note
This skill is listed under "Solutions" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is boiling point elevation and freezing point depression in?
"Solutions", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.