Solutions

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 1.

Why does a cold drink hiss when you open it?

Twist the cap and the bottle hisses, and bubbles stream up. Leave the same drink in the sun and it goes flat much sooner. A diver who surfaces too fast gets cramps that can be fatal. Salt thrown on an icy road makes the ice melt even though the air stays freezing.

All four stories come from one idea: what happens to a substance when it dissolves in something else. A solution is not just a mix. The dissolved particles change the pressure of the vapour above it, the temperature at which it boils and freezes, and the way water moves across a membrane. This chapter gives you the numbers behind each effect, and the numbers are what CUET asks for.

What is a solution?

A solution is a homogeneous mixture of two or more components. The component present in the largest amount is the solvent, and it decides the physical state of the solution. The others are solutes. We study binary solutions, with two components.

Solute and solvent can each be a gas, a liquid or a solid, which gives nine types.

Measuring how much solute: concentration units

Every unit answers the same question in a different way. The key difference is whether the denominator involves volume, which changes with temperature, or mass, which does not.

Why molarity changes with temperature. Warm a solution and the liquid expands, so the same number of moles now sits in a bigger volume, and M falls. Mass does not change on heating, which is why mass percentage, ppm, mole fraction and molality are all temperature independent. Molality is the right choice when you need an exact value over a temperature range, as in boiling and freezing point work.

Worked example: from percentage to molarity and molality. Concentrated sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.84 g mL⁻¹ (molar mass 98 g mol⁻¹). Find its molarity and molality.

The common wrong step: using 100 g (or 100 mL) as the solvent for molality. Molality needs the mass of solvent only, which here is just 2 g. That is also why molality is so much larger than molarity for concentrated solutions.

Solubility: how much dissolves, and why

Solubility is the maximum amount of a substance that dissolves in a given amount of solvent at a stated temperature (and pressure, for gases). The rule of thumb is like dissolves like. Polar solutes (salts, sugar) dissolve in polar solvents (water) because new solute–solvent attractions make up for the ones broken. Non-polar solutes (naphthalene, fats) dissolve in non-polar solvents (benzene, hexane) because the weak forces on both sides are similar.

Solids in liquids. Dissolving reaches a dynamic equilibrium: Solute + Solvent ⇌ Solution. Solute particles leave the crystal and rejoin it at the same rate. The concentration of that saturated solution is the solubility. Temperature can push the equilibrium either way, depending on the enthalpy of solution.

Pressure has almost no effect on the solubility of a solid or liquid in a liquid, because these are barely compressible.

Gases in liquids: Henry's law. Push down on the gas above a liquid and more gas molecules strike the surface and dissolve, until a new equilibrium is reached. Henry's law says that, at constant temperature, the partial pressure of a gas above the solution is proportional to its mole fraction in the solution:

$$p = K_H\, x$$

where $K_H$ is Henry's law constant (unit: bar or kbar).

Read it carefully: for the same pressure, a larger $K_H$ means a smaller x. So higher $K_H$ means lower solubility.

Why $K_H$ rises with temperature. Dissolving a gas in a liquid releases heat, so by Le Chatelier's principle heating the solution pushes gas back out. A warm solution holds less gas. That is why a drink goes flat in the sun and why fish struggle in warm water, which holds less dissolved oxygen.

Worked example: nitrogen in water. Nitrogen at a partial pressure of 0.987 bar is bubbled through water at 293 K ($K_H$ = 76.48 kbar). Find the mole fraction of N₂ in solution, and the number of moles dissolved in 1 L of water.

The common wrong step: leaving $K_H$ in kbar while p is in bar, which makes x a thousand times too large.

Where Henry's law shows up. Soft drinks are sealed with CO₂ at high pressure, so opening the bottle drops the pressure and the dissolved gas escapes as bubbles. A diver breathes air at high pressure, so a lot of N₂ dissolves in the blood. If the diver surfaces too fast, the pressure falls suddenly and N₂ forms bubbles that block capillaries, causing the bends. Divers' tanks use helium diluted air, because helium is much less soluble than nitrogen. At high altitude the partial pressure of O₂ is low, so less dissolves in blood and climbers feel weak and cannot think clearly (anoxia).

Pause and predict: a sealed bottle of soda is at 5 °C. Another identical bottle is at 30 °C. Which will fizz harder on opening, and why?

Check your answer: the warm one. At the higher temperature $K_H$ is larger, so less CO₂ stays dissolved at equilibrium and more is ready to escape the moment the pressure drops. The cold bottle holds the gas more calmly.

Beyond the textbook. For the exam: $p = K_H x$ for any gas in a liquid. In reality: Henry's law works well for gases that do not react with the solvent and at modest pressures. Gases such as CO₂, NH₃ and HCl react with water, so they dissolve more than the law predicts. Why it matters: the numbers you calculate in the exam are for gases like N₂, O₂ and He that simply dissolve.

Vapour pressure and Raoult's law

Every liquid has a tendency to evaporate. In a closed container, vapour molecules go back into the liquid at the same rate that liquid molecules escape, and the pressure of the vapour at that point is the vapour pressure. A more volatile liquid has a higher vapour pressure at the same temperature.

Two volatile liquids. When two liquids 1 and 2 are mixed, each contributes to the vapour. Raoult's law says the partial vapour pressure of each volatile component is proportional to its mole fraction in the liquid:

$$p_1 = x_1\, p_1^\circ \qquad p_2 = x_2\, p_2^\circ$$

where $p^\circ$ is the vapour pressure of the pure liquid.

By Dalton's law the total pressure is their sum:

$$p_{\text{total}} = p_1^\circ + (p_2^\circ - p_1^\circ)\, x_2$$

The vapour has a different composition from the liquid. The mole fraction of component i in the vapour is

$$y_i = \dfrac{p_i}{p_{\text{total}}}$$

and the vapour is always richer in the more volatile component.

Worked example: a mixture of two liquids. Liquids A and B have vapour pressures 120 and 300 mm Hg at a given temperature. A solution has x(B) = 0.4. Find the total pressure and the mole fraction of B in the vapour.

B is more volatile, and the vapour has 0.625 of B against only 0.4 in the liquid. That is the principle of fractional distillation.

The common wrong step: assuming the vapour has the same composition as the liquid. It never does for a mixture of volatile liquids.

A non-volatile solute. If the solute cannot evaporate, only the solvent contributes to the vapour pressure, $p_1 = x_1\, p_1^\circ$. The solute particles occupy part of the surface, so fewer solvent molecules can escape, and the vapour pressure falls. The relative lowering of vapour pressure equals the mole fraction of the solute:

$$\dfrac{p_1^\circ - p_1}{p_1^\circ} = x_2 = \dfrac{n_2}{n_1 + n_2} \approx \dfrac{n_2}{n_1} = \dfrac{w_2 M_1}{M_2 w_1}$$

The last two forms hold for a dilute solution.

Worked example: relative lowering. 6 g of urea (60 g mol⁻¹) is dissolved in 180 g of water. The vapour pressure of pure water is 23.8 mm Hg at 298 K. Find the vapour pressure of the solution.

Raoult's law is really a special case of Henry's law, where $K_H$ equals the vapour pressure of the pure component, $p_1^\circ$.

Ideal and non-ideal solutions

An ideal solution obeys Raoult's law at every concentration. For that to happen, the attractions between unlike molecules (A–B) must be about the same as those between like molecules (A–A and B–B). Then mixing changes nothing: $\Delta_{\text{mix}}H = 0$ and $\Delta_{\text{mix}}V = 0$. Examples are n-hexane + n-heptane, bromoethane + chloroethane, and benzene + toluene.

Real mixtures usually deviate.

Why ethanol + acetone deviates positively. Pure ethanol is held together by hydrogen bonds. Acetone molecules push in between and break some of them, but cannot make equally strong bonds in return. The net attraction falls, so molecules escape more readily and the vapour pressure rises.

Why chloroform + acetone deviates negatively. The hydrogen on CHCl₃ bonds to the oxygen of acetone. The new hydrogen bond makes the mixture hold together better than the pure liquids, so vapour pressure drops.

Azeotropes. A few mixtures reach a composition where the liquid and the vapour have the same composition. They boil at a constant temperature as if they were a pure liquid, and cannot be separated by fractional distillation. Large positive deviation gives a minimum boiling azeotrope, because the vapour pressure is high so the mixture boils at a lower temperature than either component. An example is ethanol and water, with about 95% ethanol by volume. Large negative deviation gives a maximum boiling azeotrope, such as 68% nitric acid and 32% water by mass, which boils at 393.5 K.

Pause and predict: you boil a solution of 80% ethanol in water and collect the vapour. Can you keep distilling until you have pure ethanol?

Check your answer: no. The mixture runs into the 95% ethanol azeotrope, and from there the vapour has the same composition as the liquid, so no further enrichment occurs. This is why making absolute alcohol needs a further step, such as chemical drying.

Colligative properties: it only matters how many

Four properties of a solution depend only on the number of solute particles, and not on what they are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. These are colligative properties. In each case the solute reduces the solvent's tendency to escape (as vapour, or as solid), and the more particles there are, the bigger the effect.

In these formulas $w_2$ and $w_1$ are the masses of solute and solvent in grams, and $K_b$ and $K_f$ are the molal elevation and depression constants, with unit K kg mol⁻¹. They depend on the solvent only.

Boiling and freezing. A liquid boils when its vapour pressure equals the atmospheric pressure (1.013 bar). A solution has a lower vapour pressure at every temperature, so it must be heated further to reach 1.013 bar, which raises the boiling point. By the same logic, the solution and the solid solvent reach equal vapour pressure only at a lower temperature, so the freezing point falls. The size of the change is directly proportional to molality.

Worked example: antifreeze. How much ethylene glycol (C₂H₆O₂, M = 62 g mol⁻¹) must be added to 4 kg of water so that it does not freeze above −6 °C? ($K_f$ = 1.86 K kg mol⁻¹)

The common wrong step: converting the temperature to kelvin as if it were a difference. ΔT is the same number in °C and in K, so use 6 directly.

Osmosis. A semipermeable membrane lets solvent molecules through but not solute. When pure solvent (or a dilute solution) is on one side and a more concentrated solution is on the other, solvent flows into the concentrated side. The extra pressure that must be applied to the solution side to stop this flow is the osmotic pressure, π. For dilute solutions $\pi = CRT$, which looks just like the ideal gas equation. Use R = 0.083 L bar K⁻¹ mol⁻¹ with C in mol L⁻¹ and the answer in bar.

Osmotic pressure is the colligative property of choice for large molecules such as proteins and polymers. It is measured at room temperature, and even a tiny concentration gives a measurable pressure, while the boiling and freezing point changes would be too small to detect.

Raw mango placed in brine shrinks as water leaves, and a limp carrot placed in water turns firm because water enters its cells. In reverse osmosis, a pressure greater than π is applied to the solution side, pushing pure solvent out through the membrane. This is how seawater is desalinated, using a membrane such as cellulose acetate.

Worked example: molar mass of a polymer. A solution of 2.0 g of a polymer in 250 mL of water has an osmotic pressure of $4.0 \times 10^{-3}$ bar at 300 K. Find the molar mass of the polymer (R = 0.083 L bar K⁻¹ mol⁻¹).

The common wrong step: leaving V in mL, which makes the answer 1000 times too small.

Beyond the textbook. For the exam: $\Delta T_f$ = $K_f$ m, so more solute always gives a lower freezing point. In reality: the equation holds only for dilute solutions. Salt on icy roads works because the freezing point of salty water is lower than 0 °C, but it stops at the eutectic point of about −21 °C for sodium chloride, after which the salt no longer melts ice. Why it matters: on very cold days other salts (such as calcium chloride) are used, and the linear equation is a model for the dilute range only.

Pause and predict: two beakers hold 0.1 m glucose and 0.1 m NaCl in water. Which freezes at the lower temperature?

Check your answer: NaCl. Glucose stays as one particle per formula unit, but NaCl gives two ions, so the solution has about twice as many particles and twice the depression.

The van't Hoff factor: when particles split or stick

The formulas above assume the solute stays as single molecules. Real solutes do not always. If the solute dissociates (NaCl → Na⁺ + Cl⁻), there are more particles than the formula mass suggests. If it associates (two ethanoic acid molecules joining into a dimer in benzene), there are fewer. The van't Hoff factor, i, corrects for this:

$$i = \dfrac{\text{observed colligative property}}{\text{calculated colligative property}} = \dfrac{\text{normal molar mass}}{\text{observed molar mass}} = \dfrac{\text{moles of particles after}}{\text{moles before}}$$

All four formulas simply gain a factor $i$:

$$\Delta T_b = i K_b m \qquad \Delta T_f = i K_f m \qquad \pi = i\,CRT \qquad \dfrac{p_1^\circ - p_1}{p_1^\circ} = i\,\dfrac{n_2}{n_1}$$

The degree of dissociation or association (α) links to i. If a formula unit splits into $n$ particles, $i = 1 + \alpha(n - 1)$. If $n$ molecules join into one, $i = 1 - \alpha\left(1 - \dfrac{1}{n}\right)$. For a dimer ($n = 2$) this becomes $i = 1 - \alpha/2$.

Worked example: electrolyte with i. 5.85 g of NaCl (58.5 g mol⁻¹) is dissolved in 500 g of water. Assuming complete dissociation, find the boiling and freezing points ($K_b$ = 0.52, $K_f$ = 1.86 K kg mol⁻¹; normal boiling point 373.15 K, freezing point 273.15 K).

The common wrong step: leaving out i, which gives half the correct change.

Worked example: association. 1.22 g of benzoic acid (122 g mol⁻¹) in 50 g of benzene lowers the freezing point by 0.768 K. $K_f$ for benzene is 5.12 K kg mol⁻¹. If benzoic acid forms dimers, what percentage is associated?

Beyond the textbook. For the exam: take i = 2 for NaCl and KCl, and i = 3 for K₂SO₄. In reality: the measured value is a little lower than the whole number, because oppositely charged ions in solution attract and partly pair up. For NaCl at 0.1 m, i is about 1.87, and it approaches 2 as the solution is diluted. MgSO₄, whose ions pair more strongly, is lower still: about 1.21 at 0.1 m and 1.82 at 0.001 m. Why it matters: the exam asks you to use the whole numbers, but the dilute limit is why.

Formula sheet

Exam traps

60-second revision

Where this chapter connects

• Before this: Equilibrium supplies Le Chatelier's principle, which explains why heating changes the solubility of solids and gases, and Chemical Bonding and Molecular Structure explains the hydrogen bonding behind every deviation from Raoult's law.
• Next: Electrochemistry takes the dissolved ions from this chapter and studies how they carry current and drive cells.
• Forward link: Thermodynamics gives you the enthalpy of solution, and Chemical Kinetics uses concentration in mol L⁻¹ just as defined here.