Is it correct to say: Confusing molarity (mol/L of solution) with molality (mol/kg of solvent) in numericals.?

Molarity uses volume of SOLUTION in litres; molality uses mass of SOLVENT in kilograms. Read units carefully before substituting.

Is it correct to say: Forgetting the van't Hoff factor i for ionic solutes like NaCl, CaCl₂, K₂SO₄.?

Any electrolyte dissociates: NaCl → 2 ions (i≈2), CaCl₂ → 3 ions (i≈3). Multiply colligative property by i.

Is it correct to say: Using ΔT_f = T_solution − T_solvent (getting a negative value)?

ΔT_f is always taken as a positive magnitude: ΔT_f = T_f(pure) − T_f(solution). Same for ΔT_b.

Is it correct to say: Applying Raoult's law form p_A = x_A p_A° when solute is non-volatile without recognising p_total = p_A.?

If solute is non-volatile, only solvent contributes vapour ⇒ p_total = x_A p_A°. Relative lowering formula (p°−p)/p° = x_B is safer.

Is it correct to say: Wrong units for Kf or Kb — writing K·mol/kg instead of K·kg/mol.?

Kf, Kb have units of K·kg·mol⁻¹ (kelvin per molal). For water: Kf = 1.86, Kb = 0.52.

Is it correct to say: Assuming osmotic pressure uses volume of solvent.?

π = CRT uses molar concentration C = n_solute / V_solution. Use total solution volume, not solvent volume.

Solutions

Concentration expressions, Raoult's law, colligative properties, van't Hoff factor — NCERT Class 12 Chemistry Ch 1

📐 10 formulas✏️ 3 examples🎯 6 practice⚖️ 5-7 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Numerical on depression of freezing point or elevation of boiling point is almost guaranteed — memorise Kf and Kb units (K·kg/mol).
→Raoult's law for volatile-volatile mixture and for non-volatile solute are two DIFFERENT forms. Read the question carefully.
→Van't Hoff factor (i) is essential when solute dissociates (NaCl, CaCl2) or associates (benzoic acid dimer). Always check the state of the solute.
→Molarity depends on temperature (volume changes), molality does NOT. Board favourite MCQ.
→Osmotic pressure numerical: use R = 0.0821 L·atm/(K·mol) and T in kelvin.

📐 Formulas(10)

Molarity (M)★ Board fav

M = \frac{n _{solute}}{V _{solution} ( L )} = \frac{w _{B} \times 1000}{M _{B} \times V ( mL )}

Symbol	Meaning
$n_{solute}$	Moles of solute (mol)
$w_{B}$	Mass of solute (g)
$M_{B}$	Molar mass of solute (g/mol)
$V$	Volume of solution (L or mL as noted)

Molality (m)★ Board fav

m = \frac{n _{solute}}{W _{solvent} ( kg )} = \frac{w _{B} \times 1000}{M _{B} \times W _{A} ( g )}

Mole Fraction

x_{A} = \frac{n _{A}}{n _{A} + n _{B}}, x_{A} + x_{B} = 1

Raoult's Law (volatile solvent, non-volatile solute)★ Board fav

\frac{p _{A}^{\circ} - p _{A}}{p _{A}^{\circ}} = x_{B}

Raoult's Law (binary volatile mixture)

p_{total} = p_{A}^{\circ} x_{A} + p_{B}^{\circ} x_{B}

Elevation of Boiling Point★ Board fav

Δ T_{b} = K_{b} \cdot m \cdot i

Symbol	Meaning
$Δ T_{b}$	T_b(solution) − T_b(pure solvent) in K
$K_{b}$	Molal elevation constant (K·kg/mol)
$m$	Molality of solute (mol/kg)
$i$	Van't Hoff factor (=1 for non-electrolyte)

Depression of Freezing Point★ Board fav

Δ T_{f} = K_{f} \cdot m \cdot i

Osmotic Pressure (van't Hoff equation)★ Board fav

π = C R T \cdot i = \frac{n}{V} R T \cdot i

Van't Hoff Factor (i)

i = \frac{observed colligative property}{calculated colligative property} = 1 + (n - 1) α

Symbol	Meaning
$i$	Van't Hoff factor (dimensionless)
$n$	Number of particles produced per formula unit
$α$	Degree of dissociation (0 to 1)

Henry's Law

p = K_{H} \cdot x

✏️ Solved Examples

1Solved Exampleeasy3 steps

18 g of glucose (M = 180 g/mol) is dissolved in 1 kg of water. Calculate the molality of the solution.

Find moles of glucose

n = \frac{18}{180} = 0.1 mol

2Solved Exampleboard4 steps

The freezing point of a solution containing 5 g of an unknown non-electrolyte in 100 g of water is −0.465 °C. Determine the molar mass of the solute. (Kf for water = 1.86 K·kg/mol)

Depression in freezing point

Δ T_{f} = 0 - (- 0.465) = 0.465 K

3Solved ExampleHOTS4 steps

A 0.1 m aqueous solution of a weak acid HA shows a freezing point depression of 0.2046 K. Calculate the degree of dissociation of the acid. (Kf(H₂O) = 1.86 K·kg/mol)

Theoretical ΔT_f assuming no dissociation (i = 1)

Δ T_{f}^{c a l c} = K_{f} \cdot m = 1.86 \times 0.1 = 0.186 K

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Confusing molarity (mol/L of solution) with molality (mol/kg of solvent) in numericals.
✓Molarity uses volume of SOLUTION in litres; molality uses mass of SOLVENT in kilograms. Read units carefully before substituting.
2
Forgetting the van't Hoff factor i for ionic solutes like NaCl, CaCl₂, K₂SO₄.
✓Any electrolyte dissociates: NaCl → 2 ions (i≈2), CaCl₂ → 3 ions (i≈3). Multiply colligative property by i.
3
Using ΔT_f = T_solution − T_solvent (getting a negative value)
✓ΔT_f is always taken as a positive magnitude: ΔT_f = T_f(pure) − T_f(solution). Same for ΔT_b.
4
Applying Raoult's law form p_A = x_A p_A° when solute is non-volatile without recognising p_total = p_A.
✓If solute is non-volatile, only solvent contributes vapour ⇒ p_total = x_A p_A°. Relative lowering formula (p°−p)/p° = x_B is safer.
5
Wrong units for Kf or Kb — writing K·mol/kg instead of K·kg/mol.
✓Kf, Kb have units of K·kg·mol⁻¹ (kelvin per molal). For water: Kf = 1.86, Kb = 0.52.
6
Assuming osmotic pressure uses volume of solvent.
✓π = CRT uses molar concentration C = n_solute / V_solution. Use total solution volume, not solvent volume.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Calculate the mole fraction of ethanol in a solution containing 46 g ethanol (M = 46) and 54 g water (M = 18).
Q2
The vapour pressure of pure water at 25 °C is 23.8 mm Hg. What is the vapour pressure of a solution containing 6 g urea (M = 60) in 90 g water?
Q3
Calculate the boiling point of a solution containing 6.5 g of a non-volatile solute (M = 130) in 250 g water. Kb(H₂O) = 0.52 K·kg/mol.
$Δ T_{b} = K_{b} m$
Q4
The osmotic pressure of a 5% (w/V) solution of cane sugar (M = 342) at 300 K. R = 0.0821 L·atm/(K·mol).
Q5
A 0.5 m aqueous solution of KCl is found to freeze at −1.80 °C. Calculate the van't Hoff factor (Kf = 1.86).
Q6
State two applications of Henry's law.

📝 Notes

Solutions

A solution is a homogeneous mixture of two or more chemically non-reacting substances. The component present in larger amount is the solvent; the smaller-amount component(s) are solutes.

Types of solutions

Classification is by physical state of solvent and solute — gas-in-liquid (aerated water), solid-in-liquid (sugar in water), liquid-in-liquid (alcohol in water), and so on. NCERT focuses on binary liquid solutions (two components).

Concentration expressions — when to use which

Molarity (M): most common in stoichiometry and titrations, but changes with temperature.
Molality (m): used for colligative properties because it does not change with temperature.
Mole fraction (x): essential in Raoult's law and gas-mixture problems.
Mass percentage / ppm: used for very dilute or industrial solutions.

Ideal vs non-ideal solutions

An ideal solution obeys Raoult's law at all concentrations, has ΔH_mix = 0 and ΔV_mix = 0 (e.g. benzene + toluene, n-hexane + n-heptane).

Non-ideal solutions deviate from Raoult's law:

Positive deviation (ΔH_mix > 0): weaker A–B interactions than A–A / B–B. Vapour pressure higher than predicted; minimum-boiling azeotrope (e.g. ethanol + water at 95.4%).
Negative deviation (ΔH_mix < 0): stronger A–B interactions (H-bonding). Vapour pressure lower than predicted; maximum-boiling azeotrope (e.g. HNO₃ + water at 68%).

Colligative properties — depend only on number of particles

Four colligative properties are examinable:

Relative lowering of vapour pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure

For electrolytes, effective particle count is scaled by the van't Hoff factor i. Osmotic pressure is often preferred to measure molar mass of macromolecules because ΔT_f and ΔT_b are too small at low concentrations.

Sanity checks

Molality is unaffected by warming the flask; molarity is not.
Salt (NaCl) lowers freezing point ~2× more than an equivalent molality of sugar because i(NaCl) ≈ 2.
Boiling point elevation is always smaller in magnitude than freezing point depression for the same solvent (Kb < Kf typically).