Why do maximum-boiling azeotropes come from negative deviation?

In negatively deviating mixtures, unlike molecules attract each other more strongly than like molecules. Escape into vapour is suppressed, so vapour pressure is depressed. The mixture must be heated to a higher temperature than either pure component to boil, producing a maximum on the boiling point curve.

Vapour Pressure Of A Solution

Q: Why does adding a non-volatile solute lower the vapour pressure of a solvent?

Solute molecules occupy part of the solvent surface without contributing to the vapour. Fewer solvent molecules escape per unit time, so the vapour pressure of the solution is lower than that of the pure solvent.

Q: What is the difference between an ideal solution and a non-ideal solution?

An ideal solution obeys Raoult's law across the entire composition range with zero enthalpy and volume change on mixing. Non-ideal solutions deviate from Raoult's law showing either positive deviation with endothermic mixing or negative deviation with exothermic mixing.

Q: Why cannot an azeotrope be separated by fractional distillation?

At the azeotropic composition, liquid and vapour phases have identical compositions. Distillation exploits differences in composition, so if they are the same no further separation is possible regardless of the number of distillation stages.

Q: Is an azeotrope a compound or a mixture?

An azeotrope is a mixture. Its composition is fixed only at a specific pressure and changes when pressure changes. Compounds have fixed compositions regardless of pressure.

Q: Why does acetone-chloroform mixture show negative deviation?

The acidic hydrogen of chloroform forms a hydrogen bond with the carbonyl oxygen of acetone, an attraction that does not exist in either pure liquid. This new attraction is stronger than the pure-liquid interactions, so vapour pressure drops below the Raoult prediction.

Q: What is Konowaloff's rule?

Konowaloff's rule states that the vapour phase is always richer in the more volatile component of the mixture. If the pure vapour pressure of A exceeds that of B, then the mole fraction of A in the vapour exceeds its mole fraction in the liquid.

Q: Can Raoult's law be derived from Henry's law?

Yes. In an ideal solution the Henry constant equals the pure vapour pressure of the component, so Henry's law p equals K H times x reduces to Raoult's law p equals x times P zero. Raoult's law is the special case of Henry's law where K H equals P zero.

ChemistrySolutionsFor NEET aspirants

The vapour pressure of a solution is the pressure exerted by its vapour when in equilibrium with the liquid. Adding a non-volatile solute always lowers it (Raoult's law), and for a mixture of two volatile liquids the total pressure equals the sum of the partial pressures: $P_{T} = x_{A} P_{A}^{\circ} + x_{B} P_{B}^{\circ}$ . Ideal solutions obey Raoult's law across all compositions; non-ideal solutions show positive (repulsive) or negative (attractive) deviations that can form azeotropes - mixtures that boil at a constant composition and cannot be separated by fractional distillation.

Key Formulas - Quick Reference

Raoult's law for volatile components: $p_{A} = x_{A} P_{A}^{\circ}$ and $p_{B} = x_{B} P_{B}^{\circ}$
Total vapour pressure (Dalton's law): $P_{T} = p_{A} + p_{B} = x_{A} P_{A}^{\circ} + x_{B} P_{B}^{\circ}$
Vapour phase composition: $y_{A} = \frac{p _{A}}{P _{T}}$ and $y_{B} = \frac{p _{B}}{P _{T}}$
Relative lowering of vapour pressure: $\frac{P _{1}^{\circ} - P _{1}}{P _{1}^{\circ}} = x_{2}$ (mole fraction of solute)
Dilute-solution form: $\frac{P _{1}^{\circ} - P _{1}}{P _{1}^{\circ}} = \frac{W _{2} \cdot M _{1}}{W _{1} \cdot M _{2}}$
Ostwald-Walker: $\frac{loss in mass of solvent}{gain in mass of CaCl _{2}} = \frac{P ^{\circ} - P}{P ^{\circ}}$
Ideal solution conditions: $Δ H_{mix} = 0$ , $Δ V_{mix} = 0$ ; obeys Raoult's law for all $x$

1. What is Vapour Pressure of a Solution

Every liquid in a closed container establishes a dynamic equilibrium between its liquid and vapour phases. The pressure of the vapour at this equilibrium is called the vapour pressure of the liquid. When another substance is dissolved in the liquid, the vapour pressure typically changes.

Relative lowering of vapour pressure is the difference between the vapour pressure of the pure solvent and that of the solution, expressed relative to the pure solvent value. For example, at 20 °C the vapour pressure of pure water is 17.54 mmHg. When an aqueous solution containing 0.010 mole fraction of ethylene glycol (a non-volatile liquid) is prepared, its vapour pressure drops to 17.36 mmHg. The lowering is $Δ P = 17.54 - 17.36 = 0.18$ mmHg.

Figure 1: Why a non-volatile solute lowers vapour pressure. In the pure solvent every surface site can evaporate. In the solution the solute occupies part of the surface and cannot itself enter the vapour, so fewer solvent molecules escape per second and equilibrium settles at

P < P^{\circ}

2. Raoult's Law

Raoult's law: at a constant temperature, the partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution.

$p_{A} = x_{A} \cdot P_{A}^{\circ}, p_{B} = x_{B} \cdot P_{B}^{\circ}$

where $P_{A}^{\circ}$ and $P_{B}^{\circ}$ are the vapour pressures of the pure liquids $A$ and $B$ .

If the vapour behaves as an ideal gas, Dalton's law of partial pressures gives the total vapour pressure:

$P_{T} = p_{A} + p_{B} = x_{A} P_{A}^{\circ} + x_{B} P_{B}^{\circ}$

Figure 2: Ideal binary solution at constant temperature. Line I is

p_{A} = x_{A} P_{A}^{\circ}

, line II is

p_{B} = x_{B} P_{B}^{\circ}

, and line III is the total

P_{T} = p_{A} + p_{B}

. All three stay straight across the entire composition range, which is precisely what defines an ideal solution.

Raoult's law as a special case of Henry's law: Henry's law $p = K_{H} \cdot x$ applies to a gas dissolved in a liquid. When we set $K_{H} = P^{\circ}$ (the vapour pressure of the pure component), we recover Raoult's law $p = x \cdot P^{\circ}$ . So Raoult's law is Henry's law with the special value $K_{H} = P^{\circ}$ , which is exactly what happens when the "solute" is chemically similar to the solvent - i.e., in an ideal solution.

Solved Example 1

The vapour pressures of ethanol and methanol are 44.5 mmHg and 88.7 mmHg respectively at some temperature. An ideal solution is formed by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution.

Solution:

Molar masses: ethanol ( $C_{2} H_{5} OH$ ) = 46, methanol ( $CH_{3} OH$ ) = 32.

Moles of ethanol $= \frac{60}{46} = 1.304$ ; moles of methanol $= \frac{40}{32} = 1.250$ .

Let $A$ = methanol, $B$ = ethanol (so $P_{A}^{\circ} = 88.7$ , $P_{B}^{\circ} = 44.5$ ).

$x_{A} = \frac{1.250}{1.250 + 1.304} = 0.489$ ; $x_{B} = 1 - 0.489 = 0.511$ .

$P_{T} = x_{A} P_{A}^{\circ} + x_{B} P_{B}^{\circ} = 0.489 \times 88.7 + 0.511 \times 44.5 = 43.4 + 22.7 = 66.1$ mmHg.

Solved Example 2

The composition of the vapour over a binary ideal solution is determined by the composition of the liquid. If

x_{A}

and

y_{A}

are the mole fractions of

A

in the liquid and vapour respectively, find the value of

x_{A}

for which

y_{A} - x_{A}

has a minimum. What is the total pressure at this composition?

Solution:

Vapour-phase mole fraction of $A$ :

$y_{A} = \frac{p _{A}}{P _{T}} = \frac{x _{A} P _{A}^{\circ}}{x _{A} P _{A}^{\circ} + ( 1 - x _{A} ) P _{B}^{\circ}} = \frac{x _{A} P _{A}^{\circ}}{P _{B}^{\circ} + ( P _{A}^{\circ} - P _{B}^{\circ} ) x _{A}}$

Let $f (x_{A}) = y_{A} - x_{A}$ . Setting $\frac{df}{d x _{A}} = 0$ and solving (algebra shown expanded):

$\frac{P _{A}^{\circ} P _{B}^{\circ}}{[ P _{B}^{\circ} + ( P _{A}^{\circ} - P _{B}^{\circ} ) x _{A} ] ^{2}} - 1 = 0$

$\Rightarrow P_{B}^{\circ} + (P_{A}^{\circ} - P_{B}^{\circ}) x_{A} = P_{A}^{\circ} P_{B}^{\circ}$

$\Rightarrow x_{A} = \frac{P _{A}^{\circ} P _{B}^{\circ} - P _{B}^{\circ}}{P _{A}^{\circ} - P _{B}^{\circ}}$

Substituting back, the total pressure at this composition is:

$P_{T} = P_{A}^{\circ} \cdot P_{B}^{\circ}$ (the geometric mean of the two pure vapour pressures)

Solved Example 3

What is the composition of the vapour in equilibrium at 30 °C with a benzene-toluene solution having mole fraction of benzene = 0.400? Given

P_{B}^{\circ} = 119

torr and

P_{T}^{\circ} = 37.0

torr (subscripts denote pure benzene B and pure toluene T).

Solution:

Partial pressures in solution: $p_{B} = 0.4 \times 119 = 47.6$ torr; $p_{T} = 0.6 \times 37 = 22.2$ torr.

Total pressure: $P = 47.6 + 22.2 = 69.8$ torr.

Vapour-phase mole fraction of benzene: $y_{B} = \frac{p _{B}}{P} = \frac{47.6}{69.8} = 0.682$ ; $y_{T} = 1 - 0.682 = 0.318$ .

The vapour is richer in benzene (the more volatile component) - $y_{B} > x_{B}$ . This is Konowaloff's rule: the vapour phase is always richer in the more volatile component.

Solved Example 4

The vapour pressure of a pure liquid

A

is 100 mmHg. When a non-volatile solute is dissolved in it, the vapour pressure of the solution becomes 80 mmHg. Calculate the mole fraction of the solute in the solution.

Solution:

Applying the relative lowering formula (which is valid for a non-volatile solute):

$\frac{P _{A}^{\circ} - P _{A}}{P _{A}^{\circ}} = x_{solute}$

$\frac{100 - 80}{100} = x_{solute} = 0.20$

Mole fraction of solvent $= 1 - 0.20 = 0.80$ .

3. Ideal Solutions

An ideal solution is one in which both components obey Raoult's law across the entire range of compositions (not just in dilute limits).

The three signatures of an ideal solution:

Raoult's law holds for all $x$ : $p_{A} = x_{A} P_{A}^{\circ}$ and $p_{B} = x_{B} P_{B}^{\circ}$
No enthalpy change on mixing: $Δ H_{mix} = 0$ (no heat absorbed or evolved)
No volume change on mixing: $Δ V_{mix} = 0$ (volumes add up)

Why: ideal behaviour occurs when the A-B, A-A, and B-B intermolecular forces are all equal in strength. Then a molecule "does not know" whether its neighbours are A or B - the escape tendency of each component is unchanged by mixing, so Raoult's law holds.

Examples of near-ideal solutions: benzene ( $C_{6} H_{6}$ ) and toluene ( $C_{6} H_{5} CH_{3}$ ); $n$ -hexane and $n$ -heptane; ethylene bromide and ethylene chloride; chlorobenzene and bromobenzene. Note that each pair is chemically similar - similar functional groups, similar polarity, similar shape.

4. Non-Ideal Solutions

Solutions that do not obey Raoult's law across the whole composition range are non-ideal. They fall into two categories based on the sign of the deviation.

(a) Positive Deviation

Total observed vapour pressure is higher than the ideal prediction. This happens when A-B interactions are weaker than A-A and B-B interactions - molecules escape more easily than in the pure liquids.

Figure 3: Positive deviation from Raoult's law. Every observed curve lies above its dashed ideal line, and

P_{T}

rises to a maximum higher than both

P_{A}^{\circ}

and

P_{B}^{\circ}

. Note that observed and ideal coincide exactly at the two pure ends. Cause: A-B forces weaker than A-A and B-B. Example: ethanol-water.

Signatures:

$P_{observed} > P_{Raoult}$ for both components
$Δ H_{mix} > 0$ (endothermic - heat absorbed as weaker bonds form)
$Δ V_{mix} > 0$ (volume expands slightly on mixing)

Examples: ethanol-water, acetone-carbon disulphide, ethanol-chloroform, cyclohexane-ethanol. Ethanol-water in particular has hydrogen bonding within pure ethanol and within pure water; on mixing, each ethanol OH is next to different chemistry, disrupting the H-bonds.

(b) Negative Deviation

Total observed vapour pressure is lower than the ideal prediction. This happens when A-B interactions are stronger than A-A and B-B - the molecules "hold on to each other" and escape less readily.

Figure 4: Negative deviation from Raoult's law. Every observed curve lies below its dashed ideal line, and

P_{T}

dips to a minimum lower than both

P_{A}^{\circ}

and

P_{B}^{\circ}

. Cause: A-B forces stronger than A-A and B-B. Example: acetone-chloroform,

HNO_{3}

-water.

Signatures:

$P_{observed} < P_{Raoult}$ for both components
$Δ H_{mix} < 0$ (exothermic - heat released as stronger bonds form)
$Δ V_{mix} < 0$ (volume contracts on mixing)

Examples: acetone-chloroform (chloroform H can hydrogen-bond to acetone C=O), $HNO_{3}$ -water, $H_{2} SO_{4}$ -water, phenol-aniline.

Figure 5: The acetone-chloroform hydrogen bond. Three chlorines pull electron density away from the chloroform

C-H

, leaving that hydrogen acidic enough to hydrogen-bond to the carbonyl oxygen of acetone. This attraction exists in neither pure liquid, so escape into the vapour becomes harder and the mixture deviates negatively.

Solved Example 5

A mixture of two liquids has

Δ H_{mix} = - 12

kJ/mol and

Δ V_{mix} = - 0.8

mL/mol. Predict whether this mixture shows positive or negative deviation from Raoult's law and give one likely example.

Solution:

$Δ H_{mix} < 0$ (exothermic) and $Δ V_{mix} < 0$ (contraction) are both signatures of negative deviation. The A-B interactions are stronger than A-A and B-B, so mixing releases heat and packs molecules more tightly.

A likely example is acetone-chloroform: the acidic hydrogen of chloroform ( $Cl_{3} C-H$ ) forms a hydrogen bond with the carbonyl oxygen of acetone ( $CH_{3} -CO-CH_{3}$ ) - an attraction that does not exist in either pure liquid.

5. Measurement of Vapour Pressure Lowering - Ostwald and Walker Apparatus

The Ostwald-Walker method is a classical way to measure the relative lowering of vapour pressure. Dry air is bubbled successively through three parts of the apparatus:

Part 1: a set of bulbs containing the solution under study.
Part 2: a set of bulbs containing the pure solvent.
Part 3: a U-tube containing anhydrous calcium chloride, which absorbs all solvent vapour.

Figure 6: Ostwald-Walker apparatus. Dry air bubbles in turn through the solution, the pure solvent, and anhydrous

CaCl_{2}

. Dividing the mass lost by the solvent bulbs by the mass gained by the

CaCl_{2}

tube gives the relative lowering

\frac{P ^{\circ} - P}{P ^{\circ}}

directly, with no pressure measurement needed.

How the mass changes work:

As dry air passes through the solution, it saturates with solvent vapour and picks up mass proportional to $P$ (the vapour pressure of the solution). So the solution bulbs lose mass $\propto P$ .
The now-partially-saturated air passes through pure solvent. It picks up additional vapour proportional to $(P^{\circ} - P)$ . So the solvent bulbs lose mass $\propto (P^{\circ} - P)$ .
The CaCl $_{2}$ tube absorbs all solvent vapour, gaining a mass proportional to $P + (P^{\circ} - P) = P^{\circ}$ .

$\frac{Loss in mass of solvent}{Gain in mass of CaCl _{2}} = \frac{P ^{\circ} - P}{P ^{\circ}} = relative lowering of vapour pressure$

From the relative lowering, and using $\frac{P ^{\circ} - P}{P ^{\circ}} = x_{solute} \approx \frac{n _{solute}}{n _{solvent}}$ , one can calculate the molar mass of an unknown non-volatile solute.

Solved Example 6

Dry air is passed through a solution containing 20 g of a non-volatile organic solute in 250 mL of water, then through pure water, and finally through a U-tube containing anhydrous CaCl

_{2}

. The solution loses 26 g and the U-tube gains 26.48 g. Calculate the molar mass of the solute.

Solution:

Loss in mass of solution $= 26$ g. Gain in CaCl $_{2}$ tube $= 26.48$ g.

Therefore loss in mass of solvent (pure water bulbs) $= 26.48 - 26 = 0.48$ g.

$\frac{loss of solvent}{gain of CaCl _{2}} = \frac{P ^{\circ} - P}{P ^{\circ}} = \frac{0.48}{26.48} = x_{solute}$

For dilute solution: $x_{solute} \approx \frac{n _{solute}}{n _{water}} = \frac{20/ M}{250/18}$

$\frac{0.48}{26.48} = \frac{20 \times 18}{M \times 250}$

$M = \frac{20 \times 18 \times 26.48}{0.48 \times 250} = 79.4$ g/mol $\approx 78$ g/mol

6. Azeotropes

An azeotrope (or "constant-boiling mixture") is a binary liquid mixture that boils at a fixed temperature and has the same composition in liquid and vapour phases. Because $x_{A} = y_{A}$ , the components of an azeotrope cannot be separated by fractional distillation.

Azeotropes arise from non-ideal behaviour that produces a maximum or minimum in the vapour pressure (and hence boiling point) curve at some intermediate composition.

Figure 7: Azeotropes. (a) Positive deviation gives a maximum in

P

and therefore a minimum in

T_{b}

: ethanol-water at about 95% ethanol boils at 78 °C. (b) Negative deviation gives a minimum in

P

and a maximum in

T_{b}

HNO_{3}

-water at about 68% acid boils at 120 °C. At the azeotrope

x_{A} = y_{A}

, so distillation cannot separate the pair.

Minimum-Boiling Azeotrope

Solutions showing positive deviation from Raoult's law have a maximum in the vapour pressure curve, which corresponds to a minimum in the boiling point curve. At the azeotrope composition, the mixture boils at a lower temperature than either pure component.

Example: ethanol-water forms a minimum-boiling azeotrope at ~95% ethanol by volume, boiling at 351.15 K (78 °C), lower than either pure ethanol (78.4 °C) or pure water (100 °C). This is why fractional distillation of a fermentation mixture cannot produce 100% ethanol - it stops at the azeotropic 95%. Absolute alcohol is made by other means (e.g., addition of benzene).

Maximum-Boiling Azeotrope

Solutions showing negative deviation have a minimum in vapour pressure, which corresponds to a maximum in boiling point. The azeotrope boils at a higher temperature than either pure component.

Example: $HNO_{3}$ -water forms a maximum-boiling azeotrope at ~68% $HNO_{3}$ by mass, boiling at 393 K (120 °C), higher than either pure $HNO_{3}$ (86 °C) or water (100 °C). This is why "concentrated nitric acid" available commercially is 68% - fractional distillation of dilute acid cannot exceed this composition.

Solved Example 7

Liquids

A

and

B

form an ideal solution. The vapour pressures of pure

A

and

B

at 100 °C are 300 and 100 mmHg respectively. Vapour above a solution containing 1 mole of

A

and 1 mole of

B

is collected and completely condensed. This condensate is then heated to 100 °C and its vapours are collected and condensed again to form liquid

L

. Find the mole fraction of

A

in the vapour above

L

Solution:

Starting solution: $x_{A} = 0.5$ , $x_{B} = 0.5$ ; $P_{T} = 0.5 (300) + 0.5 (100) = 200$ mmHg.

$y_{A}$ in first vapour $= \frac{p _{A}}{P _{T}} = \frac{0.5 \times 300}{200} = 0.75$ ; $y_{B} = 0.25$ .

First condensate: now $x_{A} = 0.75$ , $x_{B} = 0.25$ ; $P_{T}^{'} = 0.75 (300) + 0.25 (100) = 250$ mmHg.

$y_{A}$ in second vapour (which becomes liquid $L$ ) $= \frac{0.75 \times 300}{250} = 0.90$ .

Liquid $L$ vapour composition: $L$ has $x_{A} = 0.90$ , $x_{B} = 0.10$ ; $P_{T}^{''} = 0.9 (300) + 0.1 (100) = 280$ mmHg.

$y_{A}$ above $L = \frac{0.90 \times 300}{280} = 0.964$ .

Notice: each successive distillation enriches the more volatile component ( $A$ ) further. This is the principle of fractional distillation.

Common Mistakes to Avoid

Watch out

Raoult's law with mole fraction of solvent gives $p_{solution} = x_{solvent} \cdot P^{\circ}$ . Do not accidentally use mole fraction of solute here.
The relative lowering formula $\frac{P ^{\circ} - P}{P ^{\circ}} = x_{solute}$ is valid only when the solute is non-volatile. For volatile solutes, use $P_{T} = x_{A} P_{A}^{\circ} + x_{B} P_{B}^{\circ}$ instead.
Vapour is always richer in the more volatile component ( $y_{A} > x_{A}$ if $P_{A}^{\circ} > P_{B}^{\circ}$ ). Do not compute $y_{A}$ as the same as $x_{A}$ .
Positive deviation $\Rightarrow$ $Δ H_{mix} > 0$ (endothermic). Negative deviation $\Rightarrow$ $Δ H_{mix} < 0$ (exothermic). Do not confuse the signs.
Azeotropes cannot be separated by fractional distillation, but they are true mixtures, not compounds. Their composition depends on external pressure.
Minimum-boiling azeotrope = maximum in $P$ (positive deviation). Maximum-boiling azeotrope = minimum in $P$ (negative deviation). "Minimum" and "maximum" swap between $T$ and $P$ .

Frequently Asked Questions

Q1. Why does adding a non-volatile solute lower the vapour pressure of a solvent?

The vapour pressure of a liquid is due to solvent molecules escaping from the surface. When a non-volatile solute is added, its molecules occupy part of the surface and do not contribute to the vapour phase. Fewer solvent molecules can escape per unit time, so the vapour pressure of the solution is lower than that of pure solvent.

Q2. What is the difference between an ideal solution and a non-ideal solution?

An ideal solution obeys Raoult's law across the entire range of compositions and has zero enthalpy and volume change on mixing. Non-ideal solutions deviate from Raoult's law and show either positive deviation (A-B interactions weaker than A-A and B-B, giving endothermic mixing) or negative deviation (A-B interactions stronger, giving exothermic mixing).

Q3. Why cannot an azeotrope be separated by fractional distillation?

At the azeotropic composition, the liquid and vapour phases have identical compositions. Distillation works by exploiting differences in composition between liquid and vapour, so if they are the same, no further separation occurs no matter how many distillation stages are used. This is why concentrated $HNO_{3}$ is stuck at 68% and ethanol-water at 95%.

Q4. Is an azeotrope a compound or a mixture?

An azeotrope is a mixture, not a compound. It has a fixed composition only at a specific pressure. If the external pressure changes, the azeotropic composition changes too. Compounds, by contrast, have fixed compositions regardless of pressure. The azeotropic ratio also does not obey the law of definite proportions in a chemical sense.

Q5. Why does acetone-chloroform mixture show negative deviation?

In pure acetone the molecules interact via weak dipole-dipole forces. In pure chloroform there are only weak dispersion forces. When mixed, the acidic hydrogen of chloroform ( $Cl_{3} C-H$ ) forms a hydrogen bond with the carbonyl oxygen of acetone (C=O). This new attraction is stronger than the interactions in either pure liquid, so molecules escape less readily - vapour pressure drops below the Raoult prediction.

Q6. What is Konowaloff's rule?

Konowaloff's rule states that at any given liquid composition, the vapour phase is always richer in the more volatile component (the one with the higher pure vapour pressure). Mathematically, if $P_{A}^{\circ} > P_{B}^{\circ}$ , then $y_{A} > x_{A}$ . This is why repeated distillation progressively enriches the more volatile species.

Q7. Can Raoult's law be derived from Henry's law?

Yes. Henry's law says $p = K_{H} x$ for a gas dissolved in a liquid. In an ideal solution, when one component acts as "solvent" and the other as "solute" but the two are chemically similar, the Henry constant $K_{H}$ becomes equal to the vapour pressure of the pure component $P^{\circ}$ . Substituting gives Raoult's law $p = x P^{\circ}$ . So Raoult's law is the special case of Henry's law where $K_{H} = P^{\circ}$ .

Q8. What is the physical meaning of a maximum-boiling azeotrope having negative deviation?

In a negatively-deviating mixture, unlike molecules attract each other more strongly than like molecules do. The molecules hold on to each other tightly, making escape into the vapour phase difficult. The vapour pressure is depressed, and since boiling occurs when vapour pressure equals atmospheric pressure, the mixture needs to be heated to a higher temperature than either pure component to boil - producing a maximum on the boiling-point diagram.

Previous year questions on Vapour Pressure Of A Solution

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