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MHT-CET Chemistry · Solutions and Colligative Properties

Van't Hoff Factor and Abnormal Molar Mass

i = (observed colligative property)/(calculated for no dissociation) = ΔTf/(Kf·m); i > 1 for dissociation, i < 1 for association, and the degree of dissociation is α = (i − 1)/(n − 1).

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Why this matters

16 PYQs, none HARD — the page that makes the electrolyte stems on every other page honest. Two-thirds are i from a measured freezing point of a 0.15–0.2 m KCl-type solution (i comes out near 1.8), the rest are the 'x K for urea, so how much for CaCl₂' ratio (3x) or AlCl₃ (4x), one percent-dissociation calculation, and the recall of which properties are colligative. Learn the ion counts and the two rearrangements; the arithmetic is one division.

Concept 1 of 3

The Van't Hoff Factor: i = ΔTf(observed)/(Kf · m), and the Ion Count for Complete Dissociation

Intuition

An electrolyte gives more particles than its formula count, so its colligative effect is larger than the molality predicts. The ratio of observed to expected is ii; a fully dissociated salt has ii equal to its number of ions, a real solution a little less.

Definition

  • i=ΔTfobsKf mi = \dfrac{\Delta T_f^{\text{obs}}}{K_f\,m}: 0.20.2 m freezing at −0.68∘-0.68^\circC gives 0.681.86×0.2=1.83\dfrac{0.68}{1.86 \times 0.2} = 1.83; 0.150.15 m at −0.51∘-0.51^\circC gives 1.821.82; 0.20.2 m at −0.7-0.7 K gives 1.881.88; 0.180.18 m at −0.54∘-0.54^\circC gives 1.611.61; 0.010.01 m at −0.056-0.056 K gives 3.003.00.
  • Complete dissociation: KCl, NaCl →2\to 2; CaCl2_2, BaCl2_2, Na2_2SO4_4 →3\to 3; AlCl3_3 →4\to 4; Al2_2(SO4_4)3_3 →5\to 5. Non-electrolytes (urea, glucose, sucrose) i=1i = 1.
  • Ratio stems: if 11 m urea gives ΔTf=x\Delta T_f = x, 11 m CaCl2_2 gives 3x3x; if a non-electrolyte gives ΔTb=x\Delta T_b = x, AlCl3_3 at the same molality gives 4x4x.
  • ΔTf=iKfm\Delta T_f = i K_f m forwards: 0.010.01 m formic acid with i=1.1i = 1.1 gives 1.1×0.01×1.86=0.0201.1 \times 0.01 \times 1.86 = 0.020 K.
  • Association (acetic acid dimerising in benzene) gives i<1i < 1 and an abnormally HIGH observed molar mass; dissociation gives an abnormally LOW one.

Van't Hoff factor

i=observedcalculated=ΔTfobsKf m=McalcMobsi = \frac{\text{observed}}{\text{calculated}} = \frac{\Delta T_f^{\text{obs}}}{K_f\,m} = \frac{M_{\text{calc}}}{M_{\text{obs}}}

Worked example

A 0.10.1 m aqueous solution of an electrolyte freezes at −0.465∘-0.465^\circC. Find ii and suggest the number of ions it gives.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Solutions and Colligative PropertiesMODERATE
0.2 molal aqueous solution of KCl freezes at −0.680∘C-0.680^\circ\text{C}. Calculate van't Hoff factor for this solution. Kf=1.86 K kg mol−1K_f=1.86\,\text{K kg mol}^{-1}

[Q84 · 10th May Shift 1 · 2023]

Using the calculated ΔTf as the observed one

Kfm=0.372K_f m = 0.372 is the EXPECTED depression for 0.20.2 m; the observed 0.680.68 goes on top. Inverting gives 0.550.55, option (D) on the −0.660-0.660 K stem.
Drill 11 more on the van't hoff factor: i = δtf(observed)/(kf · m), and the ion count for complete dissociation

Concept 2 of 3

Degree of Dissociation From i: α = (i − 1)/(n − 1)

Intuition

If a fraction α\alpha of the formula units split into nn ions, one mole becomes 1−α+nα=1+(n−1)α1 - \alpha + n\alpha = 1 + (n - 1)\alpha moles of particles. That is ii, so α=i−1n−1\alpha = \dfrac{i - 1}{n - 1}. For association into nn-mers, i=1−α+αni = 1 - \alpha + \dfrac{\alpha}{n}.

Definition

  • 0.020.02 m, ΔTf=0.046\Delta T_f = 0.046 K, n=2n = 2: i=0.0460.0372=1.236i = \dfrac{0.046}{0.0372} = 1.236, α=0.236\alpha = 0.236, i.e. 23.6%23.6\% dissociation.
  • KCl with i=1.83i = 1.83: α=0.83\alpha = 0.83. A salt of three ions with i=2.5i = 2.5: α=0.75\alpha = 0.75.
  • Observed molar mass =Mcalci= \dfrac{M_{\text{calc}}}{i}: smaller than the formula mass for dissociation, larger for association.

Degree of dissociation

i=1+(n−1)α ⇒ α=i−1n−1association: i=1−α+αni = 1 + (n - 1)\alpha \ \Rightarrow\ \alpha = \frac{i - 1}{n - 1} \qquad \text{association: } i = 1 - \alpha + \frac{\alpha}{n}

Worked example

A 0.10.1 m solution of a salt MX2_2 freezes at −0.4092∘-0.4092^\circC. Find its percent dissociation.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Solutions and Colligative PropertiesMODERATE
Calculate the percent dissociation of 0.02 m solution if its freezing point depression is 0.046 K .[Kf\left\lbrack K_{f} \right. for water  =1.86 K kg mol−1;n=2]\left. \ = 1.86\text{ }K\text{ }kg{\text{ }mol}^{- 1};n = 2 \right\rbrack

[Q54 · 19 April Shift I · 2025]

Dividing by n instead of n − 1

α=i−1n−1\alpha = \dfrac{i - 1}{n - 1}; for n=2n = 2 that is simply i−1i - 1. Dividing 0.2360.236 by 22 gives 11.8%11.8\%, close to option (A).

Concept 3 of 3

Which Properties Are Colligative, and the Statements the Exam Tests

Intuition

A colligative property depends only on the NUMBER of solute particles, not their identity. There are exactly four; a boiling point on its own is a property of the pure liquid and is not one of them.

Definition

  • The four: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure.
  • NOT colligative: boiling point, freezing point, vapour pressure themselves (they are the solvent's properties), osmosis (a process, not a property), density, viscosity.
  • True statements: a non-volatile solute LOWERS the vapour pressure and RAISES the boiling point; 0.10.1 M NaCl has a HIGHER osmotic pressure than 0.10.1 M sucrose (i≈2i \approx 2); ΔTf=Kf\Delta T_f = K_f numerically for a 11 molal solution.
PropertyColligative?Formula
Relative lowering of vapour pressureYesP∘−PP∘=x2\dfrac{P^\circ - P}{P^\circ} = x_2
Elevation of boiling pointYesΔTb=iKbm\Delta T_b = i K_b m
Depression of freezing pointYesΔTf=iKfm\Delta T_f = i K_f m
Osmotic pressureYesπ=iCRT\pi = i C R T
Boiling point (of the solution)NoAn intensive property of the liquid; its CHANGE is colligative
'Boiling point' alone is the standard wrong answer to 'which is not colligative'.
OsmosisNoA process; osmotic PRESSURE is the property
'Osmosis is a colligative property' is a planted false statement.
Four properties, all proportional to particle count; the quantities they change are not themselves colligative.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Solutions and Colligative PropertiesMODERATE
Which among the following is not colligative property?

[Q70 · 14th May Shift 1 · 2024]

Marking 'boiling point elevation' as the non-colligative one

The option list mixes 'boiling point' with 'freezing point depression'. The bare property is the odd one out; anything with 'elevation', 'depression' or 'lowering' in its name is colligative.
Drill 2 more on which properties are colligative, and the statements the exam tests

Summary — formulas & gotchas at a glance

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Formulas (2)

Reference tables (1)

Which Properties Are Colligative, and the Statements the Exam Tests6 rows
PropertyColligative?Formula
Relative lowering of vapour pressureYesP∘−PP∘=x2\dfrac{P^\circ - P}{P^\circ} = x_2
Elevation of boiling pointYesΔTb=iKbm\Delta T_b = i K_b m
Depression of freezing pointYesΔTf=iKfm\Delta T_f = i K_f m
Osmotic pressureYesπ=iCRT\pi = i C R T
Boiling point (of the solution)NoAn intensive property of the liquid; its CHANGE is colligative
'Boiling point' alone is the standard wrong answer to 'which is not colligative'.
OsmosisNoA process; osmotic PRESSURE is the property
'Osmosis is a colligative property' is a planted false statement.
Four properties, all proportional to particle count; the quantities they change are not themselves colligative.

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