Why does 20 mL of ethanol mixed with 40 mL of water not give 60 mL of solution?

Strong hydrogen bonding between ethanol OH groups and water molecules pulls the molecules closer together than they were in their pure states, causing the total volume to contract by about 3 percent.

Types Concentration And Solubility of Solutions

Q: Why is molality preferred over molarity for colligative property calculations?

Colligative properties depend on the number of solute particles per unit mass of solvent, which does not change with temperature. Molality is a mass ratio and is temperature-independent, unlike molarity which involves volume that expands or contracts with temperature.

Q: What is the difference between saturated and supersaturated solutions?

A saturated solution is in dynamic equilibrium with undissolved solute and holds the maximum dissolvable amount at a given temperature. A supersaturated solution contains more solute than the saturation limit, achieved by careful cooling of a hot saturated solution. It is metastable and crystallises on any disturbance.

Q: Why does gas solubility decrease with increasing temperature?

Dissolution of a gas in a liquid is generally exothermic. By Le Chatelier's principle, raising the temperature shifts the dissolution equilibrium backwards, releasing dissolved gas. This is why warm soda goes flat and why fish gasp in warm water.

Q: Why do deep-sea divers breathe helium-oxygen instead of ordinary air?

Under high pressure at depth, nitrogen from air dissolves excessively in blood plasma. On rapid ascent, dissolved nitrogen forms bubbles in blood causing decompression sickness (the bends). Helium is far less soluble than nitrogen in blood (higher Henry constant), so it does not cause this dangerous problem.

Q: Is Raoult's law a special case of Henry's law?

Yes. Raoult's law is a special case of Henry's law where the Henry constant equals the vapour pressure of the pure component. This holds when the solute is chemically similar to the solvent so intermolecular forces are the same for like and unlike molecule pairs, which is the definition of an ideal solution.

Q: What is the difference between substitutional and interstitial solid solutions?

In substitutional solid solutions, solute atoms replace host lattice atoms directly and must be similar in size, as in brass and steel. In interstitial solid solutions, small solute atoms occupy empty voids between larger host atoms without displacing them, as in tungsten carbide where carbon atoms fill octahedral holes in the tungsten lattice.

ChemistrySolutionsFor JEE aspirants

A solution is a homogeneous mixture where one substance, the solute, is dissolved in another, the solvent. This concept covers the three families of solutions (gaseous, liquid, and solid) with everyday examples; the six practical concentration units used across JEE and NEET numericals - molarity, molality, normality, mole fraction, mass percentage, and parts per million; the temperature and pressure dependence of gas solubility captured by Henry's law ( $p = K_{H} \cdot x$ ); and the substitutional and interstitial types of solid solutions such as brass and tungsten carbide.

Key Formulas - Quick Reference

Molarity: $M = \frac{moles of solute}{volume of solution (L)}$
Molality: $m = \frac{moles of solute}{mass of solvent (kg)}$
Normality: $N = \frac{equivalents of solute}{volume of solution (L)}$
Mole fraction of A: $x_{A} = \frac{n _{A}}{n _{A} + n _{B}}$ , and $x_{A} + x_{B} = 1$
Mass %: $\frac{w _{solute}}{w _{solution}} \times 100$ ; Volume %: $\frac{V _{solute}}{V _{solution}} \times 100$ ; Mass/Volume %: $\frac{w _{solute} ( g )}{V _{solution} ( m L )} \times 100$
Parts per million: $ppm = \frac{w _{solute}}{w _{solution}} \times 1 0^{6}$
Henry's law: $p = K_{H} \cdot x$ (partial pressure of gas $=$ Henry's constant $\times$ mole fraction in solution)
Molarity $\leftrightarrow$ Molality: $m = \frac{1000 \cdot M}{1000 d - M \cdot M _{solute}}$ , where $d$ is solution density in g/mL

1. What is a Solution

A solution is a homogeneous mixture of two or more substances present in a single phase. In every solution the majority component is called the solvent and the minority component the solute. When both components are liquids and present in comparable amounts, the distinction can be arbitrary. In this chapter we deal almost entirely with binary solutions (two components), because ternary and higher solutions rarely appear at the JEE/NEET level.

2. Different Types of Solutions

Both the solute and the solvent can independently be gas, liquid, or solid. This gives nine possible combinations, grouped into three families by the physical state of the solvent (the major component).

Type of Solution	Common Example
Gaseous Solutions (solvent is a gas)
Gas in gas	Air - a mixture of $O_{2}$ , $N_{2}$ and other gases
Liquid in gas	Chloroform vapour mixed with nitrogen gas; humid air (water vapour in air)
Solid in gas	Camphor or naphthalene in air; smoke
Liquid Solutions (solvent is a liquid)
Gas in liquid	Oxygen dissolved in water (aquatic life); $CO_{2}$ in soda
Liquid in liquid	Ethanol in water; petrol (mixture of hydrocarbons)
Solid in liquid	Sucrose in water; salt in water
Solid Solutions (solvent is a solid)
Gas in solid	Hydrogen absorbed in palladium
Liquid in solid	Amalgam of mercury with sodium; dental amalgam
Solid in solid	Alloys - brass (Cu-Zn), bronze (Cu-Sn), steel (Fe-C), gold-copper

3. Concentration Units

The concentration of a solution tells us how much solute is present in a specified amount of solvent or solution. Because the amount can be measured by volume, mass, or moles, several concentration units coexist. The one you should choose depends on what stays constant in the problem: molarity is temperature-dependent (volume changes with $T$ ), molality is not (mass does not).

(a) Molarity (M)

Molarity is the number of moles of solute present in one litre of solution.

$M = \frac{moles of solute}{volume of solution in litres}$

Unit: mol L $^{- 1}$ (also written as M). Depends on temperature because the volume of the solution expands or contracts with temperature.

(b) Molality (m)

Molality is the number of moles of solute per one kilogram of solvent (not solution).

$m = \frac{moles of solute}{mass of solvent in kilograms}$

Unit: mol kg $^{- 1}$ (also written as m). Independent of temperature - mass does not change with $T$ . Preferred for colligative-property calculations.

Figure 1: Molarity versus molality. Molarity is measured against the volume of the whole solution, which expands on heating, so

M

changes with temperature. Molality is measured against the mass of solvent alone, weighed before the solute goes in, so

m

is temperature independent. This is why colligative-property formulas are written in molality.

(c) Normality (N)

Normality is the number of gram-equivalents of solute per litre of solution.

$N = \frac{equivalents of solute}{volume of solution in litres}$

Relation with molarity: $N = M \times n$ , where $n$ is the $n$ -factor (valency factor - number of $H^{+}$ or $OH^{-}$ or electrons transferred, depending on the reaction context).

(d) Mole Fraction

Mole fraction of a component is the ratio of its moles to the total moles in the solution.

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

Always $x_{A} + x_{B} = 1$ (for a binary solution). Dimensionless. Independent of temperature.

(e) Mass Percentage (% w/w)

$Mass % = \frac{mass of solute}{mass of solution} \times 100$

Concentrated $HCl$ is labelled as 37% w/w - meaning 37 g of $HCl$ in every 100 g of solution.

(f) Volume Percentage and Mass/Volume Percentage

Volume %: $\frac{V _{solute}}{V _{solution}} \times 100$ Mass/Volume %: $\frac{w _{solute} ( g )}{V _{solution} ( m L )} \times 100$

Volume % is common for liquid-liquid mixtures (35% v/v ethanol in water). Mass/volume % is common in medicine and biology (a 0.9% w/v saline drip = 0.9 g NaCl in 100 mL saline).

(g) Parts per Million (ppm)

Used when the solute is in extremely low concentration - dissolved oxygen in water, pollutants in air, trace ions in drinking water.

$ppm = \frac{mass of solute}{mass of solution} \times 1 0^{6}$

1 ppm means 1 part of solute per million parts of solution. For dilute aqueous solutions ( $d \approx 1$ g/mL), 1 ppm $\approx$ 1 mg/L.

Solved Example 1

5 g of NaCl is dissolved in 1000 g of water. If the density of the resulting solution is 0.997 g/cm

^{3}

, calculate the molality, molarity, normality, and mole fraction of the solute.

Solution:

Moles of NaCl $= \frac{5}{58.5} = 0.0854$ mol (molar mass of NaCl = 58.5 g/mol)

Molality $= \frac{moles of solute}{mass of solvent (kg)} = \frac{0.0854}{1.000} = 0.0854 m$

Volume of solution $= \frac{mass of solution}{density} = \frac{1000 + 5}{0.997} = 1008 cm^{3} = 1.008$ L

Molarity $= \frac{0.0854}{1.008} = 0.0847 M$

Normality $= 0.0847$ N (for NaCl, equivalent mass $=$ molar mass, so $M = N$ )

Mole fraction: moles of H $_{2}$ O $= \frac{1000}{18} = 55.55$

$x_{NaCl} = \frac{0.0854}{0.0854 + 55.55} = 1.535 \times 1 0^{- 3}$

Solved Example 2

20 mL of ethanol (density 0.7893 g/mL) is mixed with 40 mL of water (density 0.9971 g/mL) at 25 °C. The final solution has density 0.9571 g/mL. Calculate (i) the percentage change in total volume on mixing, and (ii) the molality of ethanol in the final solution.

Solution:

Mass of ethanol $= 20 \times 0.7893 = 15.786$ g

Mass of water $= 40 \times 0.9971 = 39.884$ g

Total mass of solution $= 15.786 + 39.884 = 55.670$ g

Total volume before mixing $= 20 + 40 = 60$ mL

Volume after mixing $= \frac{55.670}{0.9571} = 58.165$ mL

% change in volume $= \frac{60 - 58.165}{60} \times 100 = 3.05%$ contraction

(Ethanol and water contract on mixing due to strong H-bonding between $-OH$ groups.)

Molality of ethanol: moles of ethanol $= \frac{15.786}{46} = 0.343$ mol (molar mass of C $_{2}$ H $_{5}$ OH = 46 g/mol)

$m = \frac{0.343}{39.884/1000} = 8.60 m$

Solved Example 3

The density of a 3 M sodium thiosulphate (

Na_{2} S_{2} O_{3}

) solution is 1.25 g/cm

^{3}

. Calculate the molality of

Na^{+}

and

S_{2} O_{3}^{2 -}

ions and the mole fraction of the salt. (Molar mass of

Na_{2} S_{2} O_{3} = 158

g/mol.)

Solution:

Consider 1 L of solution. Mass of solution $= 1000 \times 1.25 = 1250$ g.

Mass of $Na_{2} S_{2} O_{3}$ in 1 L $= 3 \times 158 = 474$ g.

Mass of water (solvent) $= 1250 - 474 = 776$ g $= 0.776$ kg.

Molality of salt $= \frac{3}{0.776} = 3.87 m$

Since $Na_{2} S_{2} O_{3} \to 2 Na^{+} + S_{2} O_{3}^{2 -}$ on full dissociation:

Molality of $Na^{+} = 2 \times 3.87 = 7.73 m$ ; molality of $S_{2} O_{3}^{2 -} = 3.87 m$ .

Mole fraction of salt: moles of water $= \frac{776}{18} = 43.11$ ; moles of salt $= 3$ .

$x_{salt} = \frac{3}{3 + 43.11} = 0.065$

Solved Example 4

Concentrated sulphuric acid is 98%

H_{2} SO_{4}

by mass and has a density of 1.84 g/mL. Calculate its molarity and molality.

Solution:

Consider 100 g of the acid solution. It contains 98 g of $H_{2} SO_{4}$ and 2 g of water.

Volume of solution $= \frac{100}{1.84} = 54.35$ mL $= 0.05435$ L.

Moles of $H_{2} SO_{4} = \frac{98}{98} = 1.00$ mol.

Molarity $= \frac{1.00}{0.05435} = 18.4 M$

Molality $= \frac{1.00}{0.002 kg} = 500 m$

Notice that for very concentrated solutions, molality and molarity differ enormously - molality shoots up because so little solvent remains.

4. Solubility

The solubility of a solute in a solvent is the maximum amount of solute that will dissolve in a given amount of solvent at a specified temperature to give a saturated solution. Three descriptive terms are used:

Unsaturated: solution contains less than the maximum dissolvable amount of solute; more can be dissolved.
Saturated: solution is in dynamic equilibrium with excess undissolved solute; no more can dissolve at that temperature.
Supersaturated: unstable state containing more solute than the saturation limit, usually produced by careful cooling of a hot saturated solution. Any disturbance triggers crystallisation.

Solubility of Solids in Liquids

Governed by two factors:

Nature of solute and solvent - "like dissolves like." Polar solutes (NaCl, sugar) dissolve in polar solvents (water). Non-polar solutes (naphthalene, iodine) dissolve in non-polar solvents (benzene, $CCl_{4}$ ).
Effect of temperature - if dissolution is endothermic ( $Δ H_{sol} > 0$ , e.g. KNO $_{3}$ ), solubility increases with temperature. If exothermic ( $Δ H_{sol} < 0$ , e.g. anhydrous $Na_{2} SO_{4}$ ), solubility decreases with temperature. This is Le Chatelier's principle applied to the dissolution equilibrium.

Effect of pressure: negligible for solids in liquids, because solids and liquids are almost incompressible.

Figure 2: Solubility of solids in water against temperature. A rising curve means dissolution is endothermic (

Δ H_{sol} > 0

), as for

KNO_{3}

NaCl

is almost flat.

Na_{2} SO_{4}

peaks at 32.4 °C and then falls, because above that temperature its dissolution becomes exothermic. Gases behave the opposite way and always become less soluble as

T

rises.

Solubility of Gases in Liquids

Gas dissolution in liquids is common in everyday life - $CO_{2}$ in soda water, $O_{2}$ in the water of lakes and rivers (essential for aquatic life), and $H_{2}$ absorbed by palladium metal. Solubility of a gas depends on the nature of the gas and solvent, the temperature, and (unlike for solids) heavily on the pressure of the gas above the liquid.

5. Henry's Law

Henry's law: at a fixed temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

If we use the mole fraction $x$ of the gas in solution as the measure of solubility:

$p = K_{H} \cdot x$

where $p$ is the partial pressure of the gas and $K_{H}$ is Henry's law constant. The higher the value of $K_{H}$ , the lower the solubility of the gas at a given pressure.

Figure 3: Henry's law. Partial pressure

p

of a gas plotted against its mole fraction

x

in solution. Each gas gives a straight line through the origin whose slope is

K_{H}

. A steeper line means a larger

K_{H}

and therefore a less soluble gas, which is the single most commonly reversed fact in this topic.

Figure 4: Pressure and gas solubility. Sealed, the high

CO_{2}

pressure above the liquid holds a large mole fraction in solution. Opening the cap drops

p

to about 1 bar, so by

p = K_{H} x

the dissolved mole fraction falls in the same ratio and the excess escapes as bubbles.

Different gases have different $K_{H}$ values in the same solvent, so $K_{H}$ is characteristic of the gas-solvent pair. For a given gas, $K_{H}$ also depends on temperature - it generally increases as temperature rises, meaning solubility decreases with increasing temperature. This is why aquatic species survive better in cold water than warm water, and why bottled soda goes flat if left open in sunlight.

Gas	T / K	$K_{H}$ / kbar
He	293	144.97
H $_{2}$	293	69.16
N $_{2}$	293	76.48
N $_{2}$	303	88.84
O $_{2}$	293	34.86
O $_{2}$	393	46.82

Raoult's law as a special case of Henry's law: in an ideal solution of two volatile liquids, the partial pressure of each component follows $p_{A} = x_{A} P_{A}^{\circ}$ (Raoult's law). If we compare with Henry's law $p_{A} = K_{H} \cdot x_{A}$ , we see that Raoult's law is just Henry's law with $K_{H} = P_{A}^{\circ}$ (the vapour pressure of the pure component). So Raoult's law is the special case where the solute is chemically similar to the solvent.

Applications of Henry's Law

Soft drinks and soda water: bottled under high $CO_{2}$ pressure to maximise dissolved gas. Opening the bottle releases the pressure and $CO_{2}$ fizzes out.
Deep-sea diving (nitrogen narcosis and the bends): at depth, high partial pressure of $N_{2}$ in normal air dissolves excessively in blood. On surfacing, dissolved $N_{2}$ comes out as bubbles ("the bends"). Divers therefore breathe a helium-oxygen mix instead of air; helium is much less soluble than nitrogen.
Oxygen transport by haemoglobin: in lungs the partial pressure of $O_{2}$ is high, so $O_{2}$ binds efficiently to haemoglobin. In tissues where $p_{O_{2}}$ is low, oxyhaemoglobin releases $O_{2}$ for cellular respiration.

Solved Example 5

N_{2}

gas is bubbled through water at 293 K, how many millimoles of

N_{2}

dissolve in 1 L of water? Assume

N_{2}

exerts a partial pressure of 0.987 bar. Given

K_{H}

for

N_{2}

at 293 K is 76.48 kbar.

Solution:

By Henry's law: $x_{N_{2}} = \frac{p _{N_{2}}}{K _{H}} = \frac{0.987 bar}{76480 bar} = 1.29 \times 1 0^{- 5}$

1 L of water contains $\frac{1000}{18} = 55.5$ moles.

Let $n$ be the moles of $N_{2}$ dissolved. Since $n ≪ 55.5$ :

$x_{N_{2}} = \frac{n}{n + 55.5} \approx \frac{n}{55.5} = 1.29 \times 1 0^{- 5}$

$n = 1.29 \times 1 0^{- 5} \times 55.5 = 7.16 \times 1 0^{- 4}$ mol $= 0.716$ mmol

Solved Example 6

The Henry's law constant of

CO_{2}

in water at 298 K is

1.67 \times 1 0^{8}

Pa. Calculate the mass of

CO_{2}

dissolved in 500 mL of soda water when packed under 2.5 atm

CO_{2}

pressure at 298 K.

Solution:

Convert pressure: $p_{CO_{2}} = 2.5 \times 101325 = 2.533 \times 1 0^{5}$ Pa.

Mole fraction of $CO_{2}$ in water:

$x_{CO_{2}} = \frac{p _{CO_{2}}}{K _{H}} = \frac{2.533 \times 1 0 ^{5}}{1.67 \times 1 0 ^{8}} = 1.52 \times 1 0^{- 3}$

Moles of water in 500 mL $\approx \frac{500}{18} = 27.78$ mol.

Moles of $CO_{2} \approx x_{CO_{2}} \times 27.78 = 1.52 \times 1 0^{- 3} \times 27.78 = 0.0422$ mol.

Mass of $CO_{2}$ $= 0.0422 \times 44 = 1.86$ g.

6. Solid Solutions

Solid solutions are formed by mixing two solid components in the molten state in the right proportions and letting the mixture cool. The atoms of one substance take up positions in the crystal lattice of the other. Two structural types exist:

Figure 5: Two types of solid solution. (a) Substitutional: solute atoms of comparable size (typically within 15%) take the place of host atoms, as in brass (Cu-Zn) and bronze (Cu-Sn). (b) Interstitial: much smaller solute atoms occupy the voids between host atoms without displacing any of them, as in tungsten carbide (WC).

Substitutional Solid Solutions

Atoms, ions, or molecules of the solute replace the corresponding species of the solvent in its crystal lattice. This requires the two species to have comparable sizes (typically within 15% of each other). Familiar examples: brass (Cu-Zn), bronze (Cu-Sn), monel metal, and various types of steel.

Interstitial Solid Solutions

Small solute atoms occupy the empty voids (interstices) between the larger host atoms in the lattice. The classic example is tungsten carbide (WC): tungsten atoms form a face-centred cubic pattern and carbon atoms sit in the octahedral holes, each surrounded by six tungsten atoms at the vertices of an octahedron. Tungsten carbide is extraordinarily hard and is used to make cutting tools and drilling equipment.

Common Mistakes to Avoid

Watch out

Molarity uses volume of solution, molality uses mass of solvent. A very common exam trap is to plug the wrong denominator. Always check what the problem asks for.
Molarity is temperature-dependent - it changes when the solution is heated because volume changes. Molality and mole fraction do not change with temperature.
1 kg of solvent $\neq =$ 1 L of solvent for anything other than water. Even for water this holds only near 4 °C where density is exactly 1 g/mL.
In Henry's law $p = K_{H} \cdot x$ , remember $x$ is the mole fraction of the gas in the liquid, not in the vapour phase.
Higher $K_{H}$ means lower solubility of the gas, not higher. This trips up many students who mistake $K_{H}$ for a solubility constant.
For dilute aqueous solutions, mole fraction of solute $x_{2} \approx \frac{n _{solute}}{n _{water}}$ . The approximation fails when the solute is comparable in moles to the solvent (concentrated solutions).

Frequently Asked Questions

Q1. Why is molality preferred over molarity for colligative property calculations?

Colligative properties depend only on the number of solute particles per unit mass of solvent, which does not change with temperature. Molality is a mass ratio and is therefore temperature-independent. Molarity involves volume, which expands or contracts with temperature, so using molarity would introduce errors when experiments are done at temperatures different from the reference.

Q2. What is the difference between saturated and supersaturated solutions?

A saturated solution is in dynamic equilibrium with excess solid solute at a given temperature; it holds exactly the maximum dissolvable amount. A supersaturated solution contains more dissolved solute than the saturation limit, achieved by carefully cooling a hot saturated solution without disturbance. It is metastable - any scratch, dust particle, or seed crystal triggers crystallisation of the excess.

Q3. Why does gas solubility decrease with increasing temperature?

Dissolution of a gas in a liquid is generally exothermic (heat is released as gas molecules are captured by solvent). By Le Chatelier's principle, raising the temperature shifts the equilibrium backwards, releasing dissolved gas. This is why warm soda goes flat quickly and why fish gasp at the surface of warm ponds.

Q4. Why do deep-sea divers breathe helium-oxygen instead of ordinary air?

Under high pressure at depth, nitrogen from ordinary air dissolves excessively in blood plasma. When a diver ascends too quickly, dissolved nitrogen comes out of solution as bubbles in the bloodstream, causing a painful and dangerous condition called decompression sickness or "the bends". Helium has a much smaller $K_{H}$ than nitrogen in blood and is far less soluble, so it does not cause this problem.

Q5. Is Raoult's law a special case of Henry's law or vice versa?

Raoult's law is a special case of Henry's law where the Henry constant $K_{H}$ equals the vapour pressure of the pure component $P^{\circ}$ . This happens when the solute is chemically similar to the solvent, so that intermolecular forces between unlike molecules match those between like molecules - the definition of an ideal solution. For a dilute non-ideal solution, the solvent obeys Raoult's law while the solute obeys Henry's law.

Q6. Why does 20 mL of ethanol mixed with 40 mL of water not give exactly 60 mL of solution?

Ethanol and water form strong hydrogen bonds between the ethanol $-OH$ group and water molecules. These attractions pull the molecules closer together than they were in their pure states, causing the total volume to contract slightly (by ~3% in this typical case). This is why volumes should be treated cautiously in solution problems - always work with masses when precision matters.

Q7. What is the difference between substitutional and interstitial solid solutions?

In substitutional solid solutions, the solute atoms replace some of the host lattice atoms directly - they need to be similar in size (typically within 15%). Examples include brass and steel. In interstitial solid solutions, small solute atoms squeeze into the empty spaces (interstices) between the larger host atoms without displacing them. Tungsten carbide, where carbon atoms fill octahedral voids in the tungsten lattice, is the classic example - and this arrangement is why WC is so hard.

Previous year questions on Types Concentration And Solubility of Solutions

21 questions from past papers, each with a step-by-step solution.

Show all 21 questions

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