Unit 1: Solutions

Link to Std 11 : In Std 11 you counted moles and used the mole concept; here the same mole counting tells us how strong a solution is.

In the figure : Whatever the states of the two parts, the result is one uniform phase. The first group of concentration terms uses only masses, volumes of the same kind or mole counts, so it does not change when the temperature changes.

Rule 1: Types of binary solutions — Solute and solvent can each be gas, liquid or solid, giving nine combinations. Air is gas in gas, soda water is gas in liquid, amalgam of mercury with sodium is liquid in solid, and brass (zinc in copper) is solid in solid. Solid in solid solutions are called alloys.

• Easy : Sugar in water is a solid in liquid solution; water is the solvent.
• Medium : Soda water: carbon dioxide is the solute and water is the solvent, so the solution is a liquid.
• Exam-type : A 22-carat gold ornament has 22 parts gold and 2 parts copper. It is a solid solution (an alloy) and gold, the larger part, acts as the solvent.

Rule 2: Mass percentage and ppm — Mass % of a component = (mass of the component ÷ total mass of solution) × 100. For very dilute solutions we use parts per million: ppm = (mass of component ÷ total mass of solution) × 10⁶.

• Easy : 12 g sugar dissolved in 88 g water gives a solution of mass 100 g, so mass % of sugar = 12 ÷ 100 × 100 = 12.0 %.
• Medium : A 250 g water sample contains 0.0020 g of dissolved fluoride: ppm = 0.0020 ÷ 250 × 10⁶ = 8.0 ppm.
• Exam-type : Volume % = (volume of solute ÷ volume of solution) × 100, so 20 mL ethanol made up to 100 mL with water is 20 % by volume; mass/volume % is grams of solute in 100 mL of solution, for example 5 g in 100 mL = 5 %.

Rule 3: Mole fraction — Mole fraction of a component = moles of that component ÷ total moles of all components. The mole fractions of all components add up to 1, and mole fraction has no unit.

• Easy : 9 g water (0.5 mol) and 46 g ethanol (1.0 mol): x(ethanol) = 1.0 ÷ 1.5 = 0.667 and x(water) = 0.333.
• Medium : In a mixture x(A) = 0.25, so x(B) = 1 − 0.25 = 0.75 for a two-component solution.
• Exam-type : 20 g NaCl in 180 g water: n(NaCl) = 20 ÷ 58.5 = 0.3419 mol, n(water) = 180 ÷ 18 = 10.0 mol, so x(NaCl) = 0.3419 ÷ 10.3419 = 0.0331.

Spot the difference

Link to Std 11 : In Std 11 you found moles from mass and used molar mass; now moles are divided by a volume or a mass to get a concentration.

In the figure : Molarity uses the volume of the whole solution; molality uses the mass of the solvent alone. A solution's volume grows when it is heated but its mass does not, so molarity changes with temperature while molality does not.

Rule 1: Calculating molarity — M = (mass of solute ÷ molar mass) ÷ volume of solution in litres. For dilution the amount of solute stays fixed, so M₁V₁ = M₂V₂.

• Easy : 5.85 g NaCl (0.100 mol) in 500 mL of solution: M = 0.100 ÷ 0.500 = 0.2 mol L⁻¹.
• Medium : 25 mL of 0.8 M solution is made up to 100 mL: M₂ = 0.8 × 25 ÷ 100 = 0.2 M.
• Exam-type : How much 12 M stock is needed to make 600 mL of 0.5 M? V₁ = 0.5 × 600 ÷ 12 = 25.0 mL.

Rule 2: Calculating molality — m = (mass of solute ÷ molar mass) ÷ mass of solvent in kg. Take the mass of the solvent, not of the solution.

• Easy : 4.0 g NaOH (0.10 mol) in 250 g water: m = 0.10 ÷ 0.250 = 0.4 mol kg⁻¹.
• Medium : 1.0 mol of glucose in 2.0 kg of water has molality 0.50 mol kg⁻¹.
• Exam-type : A 10 % by mass solution of glucose (molar mass 180): 10 g glucose is in 90 g water, so m = (10 ÷ 180) ÷ 0.090 = 0.617 mol kg⁻¹.

Rule 3: Changing between units — Pick a convenient basis such as 100 g of solution, find moles of solute and mass of solvent for molality, and use density to get the volume for molarity.

• Easy : 98 % H₂SO₄ by mass: in 100 g there are 98 g acid = 1.0 mol and 2 g water.
• Medium : Density 1.84 g mL⁻¹, so 100 g occupies 100 ÷ 1.84 = 54.35 mL = 0.05435 L.
• Exam-type : M = 1.0 ÷ 0.05435 = 18.4 mol L⁻¹ and m = 1.0 ÷ 0.002 = 500 mol kg⁻¹; the two values are very different for such a concentrated solution.

Spot the difference

Link to Std 11 : In Std 11 you learnt that equilibrium is dynamic and that Le Chatelier's principle predicts shifts; both are used here.

In the figure : The steeper line belongs to the gas with the larger KH: it needs more pressure to dissolve the same mole fraction, so it is less soluble. Doubling the pressure doubles the dissolved amount along each line.

Rule 1: Solids in liquids — "Like dissolves like": polar and ionic solids dissolve in polar solvents such as water, non-polar solids in non-polar solvents. In a saturated solution dissolving and crystallising go on at the same rate (dynamic equilibrium). If dissolving absorbs heat, solubility rises with temperature; if it releases heat, solubility falls. Pressure has almost no effect on solids.

• Easy : Sugar dissolves in water but not in kerosene; naphthalene dissolves in benzene but not in water.
• Medium : Solubility of KNO₃ rises sharply on heating because its dissolving is endothermic.
• Exam-type : A saturated solution is cooled and crystals appear: the solubility fell, so the dissolving process was endothermic.

Rule 2: Gases in liquids: Henry's law — p = KH x. A larger KH means lower solubility at the same pressure. At a fixed pressure, solubility of a gas falls as the temperature rises, because dissolving is exothermic and KH increases with temperature.

• Easy : Gas X with KH = 5.0 × 10⁴ bar at 2.0 bar: x = 2.0 ÷ 50000 = 4.0 × 10⁻⁵.
• Medium : In 1 L of water (55.6 mol) this is about 2.22 × 10⁻³ mol of X.
• Exam-type : If the pressure is raised from 2.0 to 6.0 bar at the same temperature, x becomes 3 times larger: 1.2 × 10⁻⁴.

Rule 3: Uses of Henry's law — Carbonated drinks are sealed under high CO₂ pressure to keep more gas dissolved. Divers breathe air at high pressure, which dissolves extra nitrogen in blood; on rapid ascent bubbles form and cause bends, so divers use helium diluted with oxygen. At high altitude low oxygen pressure gives anoxia: weakness and poor thinking in climbers.

• Easy : Fizzy drink bottles are filled at high pressure of CO₂.
• Medium : Helium is used for divers because it is much less soluble in blood than nitrogen.
• Exam-type : Why do fish die in warm ponds in summer? The solubility of oxygen falls with rising temperature, so less oxygen reaches them.

Spot the difference

Link to Std 11 : In Std 11 you met vapour pressure and intermolecular forces; this section links them to the composition of a mixture.

In the figure : The straight line is the ideal solution. If A–B attractions are weaker than A–A and B–B, molecules escape easily and the curve lies above the line; if A–B attractions are stronger, the curve lies below it.

Rule 1: Two volatile liquids — p(A) = x(A) p°(A), p(B) = x(B) p°(B), so p(total) = p°(A) + [p°(B) − p°(A)] x(B). The vapour is richer in the more volatile component: mole fraction in vapour y(A) = p(A) ÷ p(total).

• Easy : A has p° = 300 mm Hg and B has 120 mm Hg; with x(A) = 0.40: p(A) = 0.40 × 300 = 120 mm Hg.
• Medium : p(B) = 0.60 × 120 = 72 mm Hg, so p(total) = 192 mm Hg.
• Exam-type : Vapour composition: y(A) = 120 ÷ 192 = 0.625, larger than x(A) = 0.40, because A is the more volatile liquid.

Rule 2: Ideal solutions — An ideal solution obeys Raoult's law at every composition and temperature. Its mixing has ΔH = 0 and ΔV = 0 because A–B forces are about the same as A–A and B–B forces. Examples: n-hexane with n-heptane, bromoethane with chloroethane.

• Easy : Mixing two very similar liquids such as n-hexane and n-heptane is close to ideal.
• Medium : Mixing is not accompanied by heat change or volume change in an ideal solution.
• Exam-type : Check: 50 mL + 50 mL of an ideal pair gives exactly 100 mL; non-ideal pairs do not.

Rule 3: Non-ideal solutions — Positive deviation: A–B forces are weaker, vapour pressure is higher than expected, ΔH > 0, ΔV > 0 (ethanol with acetone). Negative deviation: A–B forces are stronger, vapour pressure is lower, ΔH < 0, ΔV < 0 (chloroform with acetone, because of hydrogen bonding).

• Easy : Ethanol + acetone: acetone breaks the hydrogen bonds of ethanol, so the mixture has a higher vapour pressure than predicted.
• Medium : Chloroform + acetone gets warm on mixing, which shows ΔH is negative.
• Exam-type : A solution has 100 mL liquid X + 100 mL liquid Y = 196 mL. The volume shrank, so ΔV < 0 and the deviation is negative.

Rule 4: Azeotropes — Azeotropes are mixtures of fixed composition that boil at a constant temperature and cannot be separated by fractional distillation. Positive deviation gives a minimum-boiling azeotrope (ethanol–water, about 95 % ethanol); negative deviation gives a maximum-boiling azeotrope (nitric acid–water, about 68 % acid).

• Easy : You cannot get 100 % ethanol from an ethanol–water mixture by simple distillation.
• Medium : Minimum-boiling azeotropes come from positive deviation; maximum-boiling ones from negative deviation.
• Exam-type : A mixture shows a boiling-point minimum at one composition: it must show positive deviation from Raoult's law.

Spot the difference

Link to Std 11 : You know that boiling occurs when vapour pressure equals atmospheric pressure; that idea explains the shifts below.

In the figure : A non-volatile solute lowers the vapour pressure, so the solution's curve lies below that of the pure solvent. It must be heated more to reach 1 atm (boiling point rises by ΔTb) and cooled more to freeze (freezing point falls by ΔTf).

Rule 1: Relative lowering of vapour pressure — (p° − p) ÷ p° = x(solute) = n(2) ÷ (n(1) + n(2)). For dilute solutions this is about n(2) ÷ n(1), and it gives molar mass: M(2) = (w(2) × M(1) × p°) ÷ (w(1) × (p° − p)).

• Easy : 9.0 g glucose (0.050 mol) in 90 g water (5.0 mol): x(glucose) = 0.050 ÷ 5.050 = 0.0099.
• Medium : If p° of the solvent is 30.0 mm Hg, the lowering is 0.0099 × 30.0 = 0.297 mm Hg, so p = 29.703 mm Hg.
• Exam-type : For equal masses of two non-electrolytes, the one with the smaller molar mass has more moles and lowers the vapour pressure more.

Rule 2: Elevation of boiling point — ΔTb = Kb × m, where Kb (K kg mol⁻¹) is the molal elevation constant of the solvent. Molar mass of the solute: M(2) = (Kb × w(2) × 1000) ÷ (ΔTb × w(1)), with w(1) the mass of solvent in g.

• Easy : 6.0 g urea (60 g mol⁻¹) in 200 g water: m = 0.10 ÷ 0.200 = 0.5 mol kg⁻¹.
• Medium : With Kb = 0.52 K kg mol⁻¹: ΔTb = 0.52 × 0.50 = 0.26 K, so the boiling point is 373.41 K.
• Exam-type : An unknown (2.0 g) in 50 g water raises the boiling point by 0.26 K: M = (0.52 × 2.0 × 1000) ÷ (0.26 × 50) = 80.0 g mol⁻¹.

Rule 3: Depression of freezing point — ΔTf = Kf × m, with Kf the molal depression constant (water: 1.86 K kg mol⁻¹). Molar mass: M(2) = (Kf × w(2) × 1000) ÷ (ΔTf × w(1)).

• Easy : The same urea solution (m = 0.50): ΔTf = 1.86 × 0.50 = 0.93 K, so it freezes at 272.22 K.
• Medium : Antifreeze (ethylene glycol) in car radiators lowers the freezing point of water.
• Exam-type : 3.0 g of an unknown in 100 g water lowers the freezing point by 0.62 K: M = (1.86 × 3.0 × 1000) ÷ (0.62 × 100) = 90.0 g mol⁻¹.

Spot the difference

Link to Std 11 : You know diffusion of particles from high to low concentration; osmosis is a special case where only the solvent can pass.

In the figure : Left: solvent moves towards the solution. Middle: the rising liquid makes a pressure that stops the net flow; that pressure is π. Right: if a pressure larger than π is applied on the solution, solvent flows out of it, which is reverse osmosis.

Rule 1: Osmotic pressure — π = (n ÷ V) R T. Molar mass of an unknown solute: M = (w R T) ÷ (π V). Solutions of the same osmotic pressure are isotonic; a solution with higher π is hypertonic, with lower π is hypotonic.

• Easy : 1.8 g glucose (0.010 mol) in 500 mL at 300 K: π = 0.020 × 0.0821 × 300 = 0.493 atm.
• Medium : Normal saline (0.9 % NaCl) is isotonic with blood, so red blood cells neither swell nor shrink in it.
• Exam-type : A cell placed in a hypertonic solution loses water and shrinks; in a hypotonic solution it swells and may burst.

Rule 2: Reverse osmosis — If a pressure greater than π is applied on the solution side, solvent moves from solution to the pure side. Used to turn sea water into drinking water (desalination) with a cellulose-acetate type membrane.

• Easy : Sea water has π of about 25 atm, so a pressure above this is needed.
• Medium : Home water purifiers use a reverse osmosis membrane.
• Exam-type : RO needs an applied pressure higher than π; ordinary osmosis needs no applied pressure.

Rule 3: Abnormal molar mass and van't Hoff factor — If the solute dissociates or associates, the particle count changes. i = (observed colligative property) ÷ (calculated value for no change) = (normal molar mass) ÷ (observed molar mass) = (total moles after) ÷ (moles before). Then ΔTf = i Kf m, ΔTb = i Kb m, π = i C R T.

• Easy : NaCl fully dissociated gives 2 particles, so i is about 2; K₂SO₄ gives 3 ions, so i is about 3.
• Medium : Ethanoic acid in benzene forms dimers, so the number of particles falls and i is about 0.5.
• Exam-type : For a solute AB that dissociates to A⁺ and B⁻ with degree α = 0.8: i = 1 + α = 1.8. For dimerisation with degree β = 0.8: i = 1 − β ÷ 2 = 0.6.

Rule 4: Using i in calculations — i can be less than 1 (association) or greater than 1 (dissociation). A larger i means a larger colligative effect at the same molality.

• Easy : A non-electrolyte like glucose has i = 1.
• Medium : Compare 0.1 m glucose and 0.1 m NaCl: the NaCl solution has about twice the ΔTf.
• Exam-type : 0.20 m MgCl₂ with i = 2.7 and Kf = 1.86: ΔTf = 2.7 × 1.86 × 0.20 = 1.004 K.

Spot the difference