Solutions ISC Class 12 Chemistry: Notes and Numericals

What Solutions is about

Weightage: Solutions is part of the Physical Chemistry group, which carries 25 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Concentration terms

Molarity (M) = moles of solute per litre of solution. Molality (m) = moles of solute per kilogram of solvent. Normality (N) = gram equivalents of solute per litre of solution; N = M × n-factor (n-factor = basicity of an acid, acidity of a base, or electrons gained or lost per formula unit). Mole fraction $x_A=\frac{n_A}{n_A+n_B}$, and the mole fractions add to 1. ppm = mass of solute per million parts of solution by mass, used for very dilute solutions such as dissolved oxygen or fluoride in water. Mass percentage = $\frac{w_B}{w_{solution}}\times 100$. Molality, mole fraction and mass percentage are temperature-independent (they use masses); molarity and normality change with temperature because the volume of solution expands on heating. That is why molality is used in boiling and freezing point formulas.

Worked example

4.9 g of sulphuric acid ($M_r$ = 98) is dissolved in water to make 500 mL of solution. Find its molarity and normality.

Try it first, then show the solutionSolution

1. Moles = 4.9/98 = 0.05 mol; molarity = 0.05/0.500 = 0.1 M
2. H₂SO₄ is dibasic (n-factor 2): equivalents = 4.9/49 = 0.1; normality = 0.1/0.500 = 0.2 N

Answer: 0.1 M and 0.2 N

Solubility of gases and Henry's law

Henry's law: at a constant temperature, the mole fraction of a gas dissolved in a liquid is proportional to the partial pressure of the gas above the liquid: $p=K_{H}x$, where $K_H$ is Henry's law constant. A larger $K_H$ means a gas is *less* soluble. Solubility of a gas falls as temperature rises (dissolution of gases is exothermic and gas molecules escape more easily), so $K_H$ increases with temperature. Applications: soft-drink bottles are sealed under high CO₂ pressure; deep-sea divers breathing compressed air get nitrogen dissolved in blood, which bubbles out on rapid ascent (the bends), so their tanks use air diluted with helium; at high altitude the low partial pressure of oxygen lowers dissolved oxygen in blood (anoxia). The law fails for gases that react with the solvent (NH₃, HCl in water) or at high pressure.

Worked example

$K_H$ for CO₂ in water at 298 K is $1.67\times 10^8$ Pa. Find the mass of CO₂ dissolved in 500 g of water when the CO₂ pressure is $2.53\times 10^5$ Pa.

Try it first, then show the solutionSolution

1. $x_{C{O}_2}=\frac{p}{K_H}=\frac{2.53\times 10^5}{1.67\times 10^8}=1.515\times 10^{-3}$
2. Moles of water = 500/18 = 27.78; since x is tiny, $n_{C{O}_2}\approx x\times n_{water}=1.515\times 10^{-3}\times 27.78=0.0421$ mol
3. Mass = 0.0421 × 44 = 1.85 g

Answer: About 1.85 g of CO₂

Raoult's law, ideal and non-ideal solutions

For a solution of two volatile liquids: the partial vapour pressure of each component equals its mole fraction times the vapour pressure of the pure component: $p_A=x_A{p}_A^0$, $p_B=x_B{p}_B^0$, and $P_{total}=p_A+p_B$. For a non-volatile solute: only the solvent contributes, so $p=x_A{p}^0$ and the relative lowering of vapour pressure equals the mole fraction of the solute: $\frac{p^0-p}{p^0}=x_B$. An ideal solution obeys Raoult's law at all compositions, with $\Delta H_{mix}=0$ and $\Delta V_{mix}=0$ (A-B attractions equal to A-A and B-B), e.g. benzene + toluene, n-hexane + n-heptane. Positive deviation: A-B attractions weaker, vapour pressure higher than predicted, $\Delta H_{mix}>0$, $\Delta V_{mix}>0$; e.g. ethanol + water, acetone + carbon disulphide. Negative deviation: A-B attractions stronger (often hydrogen bonding), vapour pressure lower, $\Delta H_{mix}<0$, $\Delta V_{mix}<0$; e.g. chloroform + acetone, nitric acid + water.

Worked example

Benzene ($p^0$ = 100 mm Hg) and toluene ($p^0$ = 40 mm Hg) form an ideal solution with mole fraction of benzene 0.4. Find the total vapour pressure and the mole fraction of benzene in the vapour.

Try it first, then show the solutionSolution

1. $p_{benzene}=0.4\times 100=40$ mm Hg; $p_{toluene}=0.6\times 40=24$ mm Hg
2. $P_{total}=64$ mm Hg
3. In the vapour, $y_{benzene}=\frac{40}{64}=0.625$: the vapour is richer in the more volatile component.

Answer: 64 mm Hg; 0.625

Azeotropic mixtures and fractional distillation

An azeotrope is a liquid mixture that boils at a constant temperature and distils with the same composition in liquid and vapour, so it cannot be separated by fractional distillation. Minimum-boiling azeotropes form from large positive deviations: the vapour-pressure curve has a maximum, so the boiling-point curve has a minimum. Example: ethanol + water, 95.6% ethanol by mass, boiling at 351.1 K (below both components). That is why fermentation ethanol cannot be concentrated beyond about 95.6% by distillation. Maximum-boiling azeotropes form from large negative deviations: e.g. nitric acid + water, about 68% HNO₃, boiling at about 393.5 K; hydrochloric acid + water, about 20.2% HCl, boiling at about 383 K. Fractional distillation separates an ideal (or near-ideal) pair because the vapour is always richer in the more volatile component; repeated vaporisation and condensation in the column gives pure components. With an azeotrope, distillation gives one pure component plus the azeotrope.

Worked example

A mixture of 50% ethanol and 50% water is fractionally distilled. What distils first, and what remains? Why can't pure ethanol be obtained?

Try it first, then show the solutionSolution

1. Ethanol + water shows positive deviation and forms a minimum-boiling azeotrope (95.6% ethanol, 351.1 K).
2. The azeotrope has the lowest boiling point, so it distils first; the excess water remains in the flask.
3. At the azeotropic composition liquid and vapour have the same composition, so further distillation cannot enrich the ethanol.

Answer: The 95.6% azeotrope distils; water remains; pure ethanol needs a non-distillation method (e.g. a dehydrating agent).

Relative lowering of vapour pressure and molar mass

A non-volatile solute occupies part of the surface, so fewer solvent molecules escape and the vapour pressure falls. By Raoult's law, $\frac{p^0-p}{p^0}=x_B=\frac{n_B}{n_A+n_B}$. For a dilute solution $n_B\ll n_A$, so $\frac{p^0-p}{p^0}\approx \frac{n_B}{n_A}=\frac{w_B{M}_A}{M_B{w}_A}$, giving $M_B=\frac{w_B{M}_A{p}^0}{w_A(p^0-p)}$. The exact form uses $\frac{p^0-p}{p}=\frac{n_B}{n_A}$. Relative lowering depends only on the mole fraction of solute, not on the solvent or temperature: it is a colligative property.

Worked example

18 g of glucose ($M_r$ = 180) is dissolved in 178.2 g of water. The vapour pressure of water at that temperature is 17.5 mm Hg. Find the vapour pressure of the solution.

Try it first, then show the solutionSolution

1. $n_{glucose}=18/180=0.1$; $n_{water}=178.2/18=9.9$
2. $x_B=\frac{0.1}{10.0}=0.01$
3. $p=p^0(1-x_B)=17.5\times 0.99=17.325$ mm Hg

Answer: 17.325 mm Hg (a lowering of 0.175 mm Hg)

Elevation of boiling point and depression of freezing point

The lowered vapour pressure means the solution must be heated to a higher temperature to boil, and cooled further before the solvent freezes. For dilute solutions $\Delta T_b=K_{b}m$ and $\Delta T_f=K_{f}m$, where $m$ is molality. Molal elevation constant $K_b$ (ebullioscopic constant): the elevation in boiling point for a 1 molal solution. Molal depression constant $K_f$ (cryoscopic constant): the depression for a 1 molal solution. Units: K kg mol⁻¹. For water $K_b$ = 0.52 and $K_f$ = 1.86 K kg mol⁻¹. Derivation (outline): from the Clausius-Clapeyron equation the lowering of vapour pressure near the boiling point is related to $\Delta T_b$ by $\frac{\Delta p}{p^0}=\frac{\Delta H_{vap}\Delta T_b}{R{T}_b^2}$. Equating with $\frac{\Delta p}{p^0}=x_B\approx \frac{m{M}_A}{1000}$ gives $\Delta T_b=\frac{R{T}_b^2{M}_A}{1000\Delta H_{vap}}m$, so $K_b=\frac{R{T}_b^2}{1000l_v}$ with $l_v$ the latent heat of vaporisation per gram. Similarly $K_f=\frac{R{T}_f^2}{1000l_f}$. Molar mass: $M_B=\frac{1000K_b{w}_B}{\Delta T_b{w}_A}$ (and the same with $K_f$, $\Delta T_f$).

Worked example

Calculate $K_f$ for water from $T_f$ = 273 K and latent heat of fusion 333.5 J g⁻¹ (R = 8.314 J K⁻¹ mol⁻¹).

Try it first, then show the solutionSolution

1. $K_f=\frac{R{T}_f^2}{1000l_f}=\frac{8.314\times 273^2}{1000\times 333.5}$
2. $=\frac{8.314\times 74529}{333500}=1.86$ K kg mol⁻¹

Answer: 1.86 K kg mol⁻¹

Osmosis, osmotic pressure and the van't Hoff laws

Osmosis is the net flow of solvent through a semipermeable membrane from pure solvent (or a dilute solution) into a more concentrated solution. Osmotic pressure $\pi$ is the excess pressure that must be applied on the solution side to just stop osmosis. For dilute solutions $\pi =CRT=\frac{n}{V}RT$, so $M_B=\frac{w_{B}RT}{\pi V}$. This is the best method for macromolecules (proteins, polymers): the effect is large and measured at room temperature. Membranes: natural (animal bladder, cell membranes, parchment) and chemical (copper ferrocyanide deposited in a porous pot). van't Hoff-Boyle's law: at constant T, $\pi \propto C$. van't Hoff-Charles' law: at constant C, $\pi \propto T$. van't Hoff-Avogadro's law: equal volumes of isotonic solutions at the same temperature contain equal numbers of solute particles. Isotonic solutions have equal $\pi$; a hypertonic solution has higher $\pi$ (more solute), a hypotonic one lower. A cell in a hypertonic solution loses water and shrinks (plasmolysis/crenation); in a hypotonic solution it swells and may burst. Reverse osmosis: applying a pressure larger than $\pi$ on the solution pushes solvent out through the membrane, used to desalinate sea water. Diffusion vs osmosis: diffusion needs no membrane and both solute and solvent move; osmosis needs a semipermeable membrane and only solvent moves.

Worked example

250 cm³ of an aqueous solution contains 1.50 g of a protein. Its osmotic pressure at 300 K is $3.20\times 10^{-3}$ atm. Find the molar mass of the protein. (R = 0.0821 L atm K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

1. Convert volume: 250 cm³ = 0.250 L
2. $M_B=\frac{w_{B}RT}{\pi V}=\frac{1.50\times 0.0821\times 300}{3.20\times 10^{-3}\times 0.250}$
3. $=\frac{36.945}{8.00\times 10^{-4}}=46181$ g mol⁻¹

Answer: About 46 200 g mol⁻¹

Abnormal molar mass and the van't Hoff factor

Colligative properties count particles. If a solute dissociates (electrolytes), there are more particles, the colligative effect is larger and the calculated molar mass is too *low*. If it associates (ethanoic or benzoic acid dimerise in benzene through hydrogen bonds), there are fewer particles and the molar mass is too *high*. van't Hoff factor: $i=\frac{\text{observed colligative property}}{\text{normal colligative property}}=\frac{\text{normal molar mass}}{\text{observed molar mass}}$. Modified formulas: $\Delta T_b=i{K}_{b}m$, $\Delta T_f=i{K}_{f}m$, $\pi =iCRT$, $\frac{p^0-p}{p^0}=i{x}_B$. Degree of dissociation (one unit gives n ions): $i=1+(n-1)\alpha$, so $\alpha =\frac{i-1}{n-1}$. Degree of association (n molecules form one): $i=1-\alpha \left(1-\frac{1}{n}\right)$, so $\alpha =\frac{1-i}{1-1/n}$; for dimerisation $\alpha =2(1-i)$. Count n carefully: NaCl 2, BaCl₂ 3, Na₂SO₄ 3, AlCl₃ 4, K₄[Fe(CN)₆] 5.

Worked example

1.22 g of benzoic acid ($M_r$ = 122) in 100 g of benzene lowers the freezing point by 0.40 K. $K_f$ for benzene = 5.12 K kg mol⁻¹. Find the degree of association of benzoic acid, which dimerises.

Try it first, then show the solutionSolution

1. Normal $\Delta T_f=K_{f}m=5.12\times \frac{1.22/122}{0.100}=0.512$ K
2. $i=\frac{0.40}{0.512}=0.781$
3. For a dimer, $i=1-\frac{\alpha }{2}$, so $\alpha =2(1-0.781)=0.4375$

Answer: About 44% associated

Chemistry in real life: Solutions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Elevation of Boiling Point and Depression of Freezing Point
• Osmosis and Osmotic Pressure
• Abnormal Molar Mass and van't Hoff Factor

Solutions formula sheet

Molarity and normality

$M=\frac{n_B}{V(L)},N=M\times n\text{-factor}$

Change with temperature.

Molality and mole fraction

$m=\frac{n_B\times 1000}{w_A(g)},x_A+x_B=1$

Independent of temperature.

Henry's law

$p=K_{H}x$

Higher $K_H$ = lower solubility.

Raoult's law (volatile components)

$P=x_A{p}_A^0+x_B{p}_B^0$

Vapour is richer in the more volatile liquid.

Relative lowering of vapour pressure

$\frac{p^0-p}{p^0}=x_B\approx \frac{w_B{M}_A}{M_B{w}_A}$

Exact: $\frac{p^0-p}{p}=\frac{n_B}{n_A}$.

Boiling point elevation

$\Delta T_b=i{K}_{b}m,M_B=\frac{1000K_b{w}_B}{\Delta T_b{w}_A}$

Water: $K_b$ = 0.52 K kg mol⁻¹.

Freezing point depression

$\Delta T_f=i{K}_{f}m,M_B=\frac{1000K_f{w}_B}{\Delta T_f{w}_A}$

Water: $K_f$ = 1.86 K kg mol⁻¹.

Ebullioscopic and cryoscopic constants

$K_b=\frac{R{T}_b^2}{1000l_v},K_f=\frac{R{T}_f^2}{1000l_f}$

$l$ = latent heat per gram (derivation included in the syllabus).

Osmotic pressure

$\pi =iCRT=\frac{i{w}_{B}RT}{M_{B}V}$

R = 0.0821 L atm K⁻¹ mol⁻¹; V in litres.

van't Hoff factor

$i=\frac{\text{normal }M}{\text{observed }M}$

i > 1 dissociation; i < 1 association.

Degree of dissociation

$\alpha =\frac{i-1}{n-1}$

n = ions from one formula unit.

Degree of association

$\alpha =\frac{1-i}{1-\frac{1}{n}}$

Dimer: $\alpha =2(1-i)$.

Solutions: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Solutions

The question patterns to expect, and the method that works for each.

Tricks to remember Solutions

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Molality loves kilograms

  molaLity: L for 'Lots of solvent mass' (per kg solvent); molaRity: R for 'Room-filling volume' (per litre of solution).

  Use it when: Choosing the right concentration term.

• Four colligatives: 'Very Big Fish Ox'

  Vapour pressure lowering, Boiling point elevation, Freezing point depression, Osmotic pressure.

  Use it when: Listing colligative properties.

• i up, M down

  Dissociation: i > 1, observed molar mass lower. Association: i < 1, observed molar mass higher.

  Use it when: Abnormal molar mass one-markers.

• Positive = Pulling apart = Plus

  Positive deviation: weaker attractions, ΔH and ΔV positive, minimum-boiling azeotrope.

  Use it when: Deviation and azeotrope questions.

• Count the ions before anything

  NaCl 2, BaCl₂ 3, Na₂SO₄ 3, AlCl₃ 4, K₄[Fe(CN)₆] 5.

  Use it when: Every van't Hoff factor numerical.

Solutions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Solutions

The slips examiners see most often, and what to write instead.

• ✗Mistake: Missing the volume conversion in osmotic pressure

  CISCE examiners' comments, 2024: in the protein molar-mass question, candidates did not convert 200 cm³ to 0.2 L before substituting, and some wrote 60.377 g/mol instead of 60,377 g/mol.

  Wrong working

  $M=\frac{1.26\times 0.0821\times 300}{2.57\times 10^{-3}\times 200}$

  Right working

  Use V = 0.200 L with R in L atm K⁻¹ mol⁻¹; write large molar masses with care (60 377, not 60.377).

  ✓Do this: Write units next to each number before substituting.

  Usually costs: 1 mark.

• ✗Mistake: Wrong number of ions (n) in van't Hoff factor problems

  CISCE examiners' comments, 2024 and 2025: candidates used the wrong n for BaCl₂, made calculation errors for K₄[Fe(CN)₆], used the wrong formula for degree of association, and in 2025 ignored i = 3 for Na₂SO₄ in an osmotic pressure calculation.

  Wrong working

  BaCl₂: n = 2; $\pi =CRT$ for Na₂SO₄.

  Right working

  BaCl₂ → Ba²⁺ + 2Cl⁻, n = 3; Na₂SO₄ → 2Na⁺ + SO₄²⁻, i = 3, $\pi =iCRT$.

  ✓Do this: Write the dissociation equation first and count ions from it.

  Usually costs: 1-2 marks.

• ✗Mistake: Getting the direction of osmosis wrong

  CISCE examiners' comments, 2024 and 2025: candidates said a shell-less egg in pure water would not change in size instead of swelling, chose hypotonic as the solution with more solute, said red blood cells in 5% NaCl would swell instead of shrink, and confused the direction of solvent flow in reverse osmosis.

  Wrong working

  RBCs in hypertonic saline swell.

  Right working

  Water leaves the cells towards the more concentrated (hypertonic) solution, so they shrink.

  ✓Do this: Ask 'which side has more solute?' and send water there.

  Usually costs: 1 mark each.

• ✗Mistake: Misreading equimolar freezing-point data

  CISCE examiners' comments, 2024: when B's freezing-point depression was half of A's, candidates chose 'normal' or 'dissociated' instead of recognising that i < 1 means association.

  Wrong working

  Smaller depression, so B is dissociated.

  Right working

  Smaller depression means fewer particles: B is associated.

  ✓Do this: Link: more particles, bigger effect; fewer particles, smaller effect.

  Usually costs: 1 mark.

• ✗Mistake: Wrong Raoult's or boiling-point formula, and vague Henry's law answers

  CISCE examiners' comments, 2025: candidates wrote an incorrect Raoult's law formula or substituted wrongly, wrote the wrong boiling-point elevation formula or miscalculated ΔTb, and could not explain that the solubility of CO₂ increases with pressure (Henry's law) for sealed soda bottles.

  Wrong working

  $\frac{p^0-p}{p^0}=x_A$ (solvent mole fraction).

  Right working

  Relative lowering equals the mole fraction of the solute, $x_B$.

  ✓Do this: Learn each formula with its symbols defined; quote Henry's law in words.

  Usually costs: 1-2 marks.

Practice questions on Solutions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which of these is independent of temperature?
   (a) molarity
   (b) normality
   (c) molality
   (d) formality

   Hint

   Which uses only masses?

   Answer

   (c) molality

2. Q2 · 1 mark · Very short answerBoard level

   Fill in the blanks: A mixture of ethanol and water shows ______ deviation from Raoult's law and forms a ______-boiling azeotrope.

   Hint

   Hydrogen bonds are broken on mixing.

   Answer

   positive; minimum

3. Q3 · 1 mark · MCQBoard level

   The van't Hoff factor of a solute that is 50% dimerised in solution is:
   (a) 0.25
   (b) 0.50
   (c) 0.75
   (d) 1.5

   Hint

   $i=1-\alpha /2$.

   Answer

   (c) 0.75

4. Q4 · 1 mark · Very short answerBoard level

   State the van't Hoff-Charles' law of osmotic pressure.

   Hint

   Constant concentration.

   Answer

   At constant concentration, the osmotic pressure of a dilute solution is directly proportional to its absolute temperature.

5. Q5 · 2 marks · Short answerHarder board level

   Why is the molar mass of NaCl found from the elevation of boiling point lower than 58.5?

   Hint

   Particles.

   Answer

   NaCl dissociates into Na⁺ and Cl⁻, nearly doubling the number of particles. The elevation is nearly doubled, and since M ∝ 1/ΔT_b the calculated molar mass is about half (around 30).

6. Q6 · 2 marks · Short answerHarder board level

   Calculate the mole fraction of ethanol in a solution containing 46 g of ethanol and 90 g of water.

   Hint

   Moles of each.

   Answer

   Ethanol 1 mol, water 5 mol: $x_{ethanol}$ = 1/6 = 0.167.

7. Q7 · 2 marks · Short answerHarder board level

   A solution of glucose in water is isotonic with a solution containing 6.0 g of urea per litre. What mass of glucose is present per litre? (urea 60, glucose 180)

   Hint

   Isotonic = same molar concentration (non-electrolytes).

   Answer

   Urea 0.1 mol L⁻¹, so glucose 0.1 mol L⁻¹ = 18 g per litre.

8. Q8 · 3 marks · Short answerHOTS / competency

   Derive the expression $K_f=\frac{R{T}_f^2}{1000l_f}$ in outline and use it to show why $K_f$ of benzene (5.12) is larger than that of water (1.86).

   Hint

   Compare latent heats.

   Answer

   From Clausius-Clapeyron and Raoult's law, $\Delta T_f=\frac{R{T}_f^2{M}_A}{1000\Delta H_{fus}}m$, i.e. $K_f=R{T}_f^2/(1000l_f)$. Benzene's latent heat of fusion per gram (about 127 J g⁻¹) is much smaller than water's (333.5 J g⁻¹), which outweighs its slightly higher $T_f$ (278.6 K), giving a larger $K_f$.

9. Q9 · 2 marks · Short answerHarder board level

   1.0 g of a non-electrolyte in 50 g of benzene lowers its freezing point by 0.40 K. Find the molar mass ($K_f$ = 5.12 K kg mol⁻¹).

   Hint

   $M=1000K_f{w}_B/(\Delta T_f{w}_A)$.

   Answer

   M = 1000 × 5.12 × 1.0/(0.40 × 50) = 256 g mol⁻¹.

10. Q10 · 3 marks · Short answerHarder board level

    An aqueous solution of an electrolyte AB₂ (0.05 mol in 500 g of water) freezes at −0.48 °C. Calculate i and the degree of dissociation. ($K_f$ = 1.86 K kg mol⁻¹)

    Hint

    AB₂ gives 3 ions.

    Answer

    m = 0.10; normal ΔT_f = 0.186 K; i = 0.48/0.186 = 2.58; α = (2.58 − 1)/2 = 0.79 (79%).

11. Q11 · 3 marks · Short answerHOTS / competency

    Two liquids A and B form an ideal solution. At 300 K, a solution with $x_A$ = 0.25 has a total vapour pressure of 550 mm Hg; one with $x_A$ = 0.50 has 600 mm Hg. Find $p_A^0$ and $p_B^0$.

    Hint

    Two simultaneous equations.

    Answer

    0.25p_A⁰ + 0.75p_B⁰ = 550 and 0.5p_A⁰ + 0.5p_B⁰ = 600. Doubling the first and subtracting the second: p_B⁰ = 500 mm Hg; then p_A⁰ = 700 mm Hg.

12. Q12 · 5 marks · Long answerHarder board level

    (a) Define osmotic pressure and state why it is preferred for finding molar masses of proteins. [2]
    (b) 2.5 g of a protein in 250 mL of solution at 27 °C has π = 2.0 mm Hg. Calculate its molar mass. [2]
    (c) What is reverse osmosis? Give one use. [1]

    Hint

    Convert mm Hg to atm (÷ 760).

    Answer

    (a) The extra pressure that must be applied to the solution to just stop osmosis of solvent into it through a semipermeable membrane. Preferred because the effect is measurable even for very dilute macromolecule solutions and is done at room temperature. (b) π = 2.0/760 = 2.63 × 10⁻³ atm; M = 2.5 × 0.0821 × 300/(2.63 × 10⁻³ × 0.250) = 93 600 g mol⁻¹. (c) Applying pressure greater than π on the solution side so that pure solvent flows out of it through the membrane; desalination of sea water.

13. Q13 · 2 marks · Short answerHOTS / competency

    Arrange in increasing order of boiling point, giving a reason: 0.1 m glucose, 0.1 m K₂SO₄, 0.1 m NaCl, 0.05 m AlCl₃.

    Hint

    Compare i × m.

    Answer

    i·m: glucose 0.1, NaCl 0.2, AlCl₃ 0.2, K₂SO₄ 0.3. Order: glucose < NaCl = AlCl₃ < K₂SO₄ (assuming complete dissociation).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define molality.Show answer

   Moles of solute per kilogram of solvent.

2. 2.State Henry's law.Show answer

   At constant temperature, the partial pressure of a gas over a solution is proportional to its mole fraction in the solution: $p=K_{H}x$.

3. 3.What is an ideal solution?Show answer

   One that obeys Raoult's law at all compositions, with $\Delta H_{mix}=0$ and $\Delta V_{mix}=0$.

4. 4.Give one example each of a minimum- and maximum-boiling azeotrope.Show answer

   Ethanol + water (95.6% ethanol); nitric acid + water (about 68% HNO₃).

5. 5.Define molal depression constant.Show answer

   The depression of freezing point produced by a 1 molal solution of a non-electrolyte.

6. 6.What are isotonic solutions?Show answer

   Solutions with the same osmotic pressure at the same temperature.

7. 7.van't Hoff factor for complete dissociation of Al₂(SO₄)₃?Show answer

   5

8. 8.Formula for degree of association for dimerisation?Show answer

   $\alpha =2(1-i)$

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you convert molarity to molality using density?
• ☐Can you state Henry's law and explain soda bottles and divers?
• ☐Can you draw vapour pressure diagrams for positive and negative deviation?
• ☐Can you name two azeotropes and their type?
• ☐Can you derive $K_b=R{T}_b^2/1000l_v$?
• ☐Can you find molar mass from any of the four colligative properties?
• ☐Can you write $i$ for association and dissociation and solve for α?
• ☐Can you predict what happens to a cell in hypotonic and hypertonic solutions?

Solutions: frequently asked questions