313_Chemistry_Eng_Lesson7.pdf). Content covers sections 7.1–7.8.When sugar or salt is added to water, it dissolves to form a solution. Solutions are central to life and industry — countless chemical reactions are carried out in solution. From the saline drip in hospitals to the electrolyte in a car battery, from the sugar in your tea to the dissolved oxygen that sustains aquatic ecosystems, solutions surround us. Study of how substances dissolve and how dissolved particles affect physical properties is therefore essential for chemistry, biology, and engineering.
This lesson covers the components of solutions, ways to express concentration, types of solutions, vapour pressure, Henry's and Raoult's laws, ideal and non-ideal behaviour, colligative properties (relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure), abnormal colligative properties, and the van't Hoff factor with numerical applications. By the end you should be able to express concentrations in multiple units, predict how mixtures behave relative to ideal Raoult's law, calculate boiling and freezing point changes, determine molar masses from colligative data, and correct for electrolyte dissociation using i.
When solute mixes homogeneously with solvent: solute + solvent → solution. A solution is a homogeneous mixture of two or more substances. The solvent is the component with the same physical state as the solution; the solute is dissolved in it. In sugar water, water is solvent and sugar is solute.
Properties like sweetness or colour depend on how much solute is present relative to solvent — the concentration. NIOS covers five main expressions:
Example 7.2: 50% H₂SO₄, density 1.20 g/cm³ → 600 g acid + 600 g water in 1 L → molality ≈ 6.8 m. Example 7.3: 36 g water + 46 g ethanol → xwater = 0.67, xethanol = 0.33. Example 7.4: 0.4 g NaOH in 100 mL → 0.1 N.
Choosing the right unit matters in exams: molarity is convenient for volumetric lab work (titrations) because solutions are measured in litres, but molality is mandatory when temperature changes during an experiment would alter the volume and hence molarity. Normality is especially useful in redox and acid–base calculations where reactions depend on equivalents rather than moles — one equivalent of acid neutralises one equivalent of base regardless of whether the acid is monobasic or dibasic.
Equivalent weight of an acid = molecular weight / basicity; of a base = molecular weight / acidity; of a salt = molecular weight / total metal valency. Oxidising and reducing agents may have different equivalent weights under different reaction conditions, calculated from the specific redox reaction they undergo.
Binary solutions can be gas–gas (air), gas–liquid (soda water), gas–solid (H₂ in Pd), liquid–gas (humidity), liquid–liquid (alcohol in water), liquid–solid (Hg in Au), solid–gas (camphor in air), solid–liquid (sugar in water), solid–solid (brass, bronze). Common cases:
Henry's law valid when: pressure not too high, temperature not too low, gas does not associate, dissociate, or react with solvent. CO₂ solubility in water: 0.88 cm³/cm³ at 20°C but 0.53 at 40°C — heating expels dissolved gas, which is why warm soda goes flat faster than cold soda.
Table 7.1 lists all nine binary solution types. In liquid–liquid solutions the component in smaller amount is usually called solute and the larger amount solvent, though this is a convention rather than a strict rule. Partial miscibility (water–phenol) produces two layers each saturated with the other component at a given temperature.
Gas solubility trends: CO₂, HCl, NH₃ are highly soluble in water; H₂, O₂, N₂ are sparingly soluble. This explains why fish need dissolved O₂ but soft drinks can hold large amounts of CO₂ under pressure.
When a pure liquid evaporates in a closed vessel, an equilibrium is reached between evaporation and condensation. The pressure exerted by the vapour is the vapour pressure of the liquid (Fig. 7.1).
For miscible volatile liquids, partial vapour pressure of each component is proportional to its mole fraction: PA = PA°·XA and PB = PB°·XB. Total pressure P = PA°·XA + PB°·XB. Solutions obeying Raoult's law at all concentrations and temperatures are ideal solutions. A plot of PA and PB versus mole fraction gives straight lines from PA° (at XA=1) to zero (at XA=0); the total pressure line connects PA° and PB° (Fig. 7.2). Raoult's law applies only when liquids are miscible — immiscible pairs form separate phases each with its own vapour pressure contribution.
The more volatile component (higher P°) contributes more to the vapour phase at any composition. Distillation of liquid mixtures exploits differences in partial vapour pressures — the vapour is enriched in the more volatile component, which is the industrial basis for separating ethanol from water and petroleum fractions.
For aqueous sugar/salt solutions, only solvent vapour is present. Since XA < 1, vapour pressure drops below pure solvent. Derivation gives:
Ideal solutions: obey Raoult's law; ΔHmix = 0 and ΔVmix = 0 — intermolecular forces between A–B are similar in strength to A–A and B–B. Benzene–toluene approximates ideal behaviour. Non-ideal solutions: most real solutions — deviate from Raoult's law with heat/volume change on mixing. Positive deviation releases heat on mixing (endothermic, ΔHmix > 0 in some cases) when breaking stronger self-interactions costs energy; negative deviation is often exothermic when new stronger A–B interactions form.
Understanding deviations helps explain azeotrope formation — mixtures that boil at constant composition because the vapour has the same composition as the liquid at a particular mole fraction, often near the maximum or minimum of the vapour pressure curve for non-ideal pairs. Ethanol–water forms an azeotrope near 95.6% ethanol, which is why simple distillation cannot produce 100% ethanol without adding a third component.
Colligative properties depend only on the number of solute particles, not their chemical nature. Four colligative properties: (1) relative lowering of vapour pressure, (2) elevation of boiling point, (3) depression of freezing point, (4) osmotic pressure. The term colligative (Latin: bound together) reflects that these effects are collectively controlled by particle count in dilute solutions.
From Section 7.5: (P°−P)/P° = XB. For dilute solutions, XB ≈ (WB/MB)/(WA/MA). Example 7.5: RLVP = 0.00715 with 7.2 g solute in 100 g water → MB ≈ 181 g mol⁻¹. VP lowering is the root cause of all other colligative effects — reduced solvent escaping tendency shifts every phase equilibrium.
Boiling point = temperature where vapour pressure equals atmospheric pressure. Solution VP < pure solvent → higher boiling point (Fig. 7.5 — solution curve lies below solvent curve, so it meets 1 atm at higher T). Derivation: ΔTb ∝ Δp ∝ XB ∝ m, giving ΔTb = Kb·m where Kb is the molal elevation constant (elevation when 1 mol non-volatile solute dissolves in 1 kg solvent). Adding salt to water raises boiling point — used in cooking (salted water boils above 100°C at 1 atm) and in antifreeze formulations.
Freezing point = temperature where solid and liquid have equal vapour pressure (Fig. 7.6). Solution VP is lower → freezing point depressed. ΔTf = Kf·m where Kf is the molal cryoscopic constant (depression when 1 mol solute in 1 kg solvent). Spreading salt on icy roads lowers the freezing point of surface water — a practical application of colligative properties. Example 7.6: 0.520 g glucose in 80.2 g water gives m = 0.036, ΔTb = 0.018 K, ΔTf = 0.66 K, so new FP = 272.34 K.
Osmosis: spontaneous flow of solvent from lower to higher concentration (or pure solvent to solution) through a semipermeable membrane. Raisins swell in water; solvent moves until concentrations equalise. Osmotic pressure (π) is the excess pressure applied to the concentrated side to stop solvent flow (Fig. 7.7). Equal π → isotonic solutions.
Osmotic pressure is preferred for macromolecules/proteins because other colligative effects are too small at low concentrations, but π remains measurable at room temperature where biomolecules are stable. Medical saline (0.9% NaCl) is approximately isotonic with blood — same osmotic pressure prevents haemolysis or cell shrinkage. Reverse osmosis applies pressure greater than π to force pure water through a membrane from seawater — used for desalination and water purification plants.
All four colligative properties are interrelated through the same thermodynamic origin: adding non-volatile solute lowers solvent chemical potential, reducing vapour pressure, which shifts boiling and freezing equilibria and creates osmotic pressure differences across membranes.
Abnormal results occur when: (i) solution is too concentrated (solute–solute interactions), (ii) association reduces effective particles (i < 1, observed M > true M — benzoic acid dimer in benzene, i ≈ ½), (iii) dissociation increases particles (i > 1, observed M < true M — NaCl in water, i ≈ 2).
Degree of association (x): 2C₆H₅COOH ⇌ dimer → at equilibrium: (1−x) mol monomer + x/2 mol dimer → effective moles = 1 − x/2. Benzoic acid i ≈ 0.5. Degree of dissociation (x): KCl ⇌ K⁺ + Cl⁻ → (1−x) + x + x = 1 + x total moles; i = 1 + x. Example 7.8: acetic acid in benzene, ΔTb gives i = 0.51, degree of association 98%. Example 7.9: 0.5% KCl freezes at 272.76 K; observed M = 38.75 vs normal 74.5 → i = 1.92, 92% dissociation.
Modified colligative equations with i: (P°−P)/P° = i·XB; ΔTb = iKbm; ΔTf = iKfm; πV = inRT. For non-electrolytes like glucose and urea, i = 1 exactly. Very concentrated solutions also show abnormal results because solute particles interact with each other, invalidating the dilute-solution assumption underlying the derivations.
Worked logic for Example 7.9 (KCl): observed M from ΔTf is half the true M because twice as many particles exist after dissociation — hence i = 74.5/38.75 = 1.92. Setting 1 + x = 1.92 gives x = 0.92 or 92% dissociated. The same logic applies to NaCl (i → 2), CaCl₂ (i → 3 if fully dissociated), and any electrolyte where each formula unit produces (1 + number of ions − 1) extra particles upon complete dissociation.
Comparison of colligative methods for molar mass: RLVP requires precise manometry; ΔTb and ΔTf need accurate thermometry but work well for small organic molecules; π is ideal for polymers and proteins (Example 7.7: M ≈ 61000 g mol⁻¹) because even dilute solutions produce measurable pressure at 300 K. Freezing point depression is often used in organic chemistry labs (cryoscopy) because many organic solvents have convenient Kf values.
Key numerical skills: convert between M and m using density; calculate mole fractions from masses; use RLVP, ΔTb, ΔTf, or π to find molar mass; compute i from observed vs normal M; find degree of dissociation from i. Water constants to remember: Kf = 1.86 K kg mol⁻¹, Kb = 0.52 K kg mol⁻¹ (textbook values). Always identify whether solute is electrolyte (i > 1) or associates (i < 1) before applying colligative formulas.
Terminal exercise covers: defining colligative properties; predicting SF₆ is outside this chapter; computing M, m, N; explaining why osmotic pressure suits proteins; calculating i and degree of dissociation for electrolytes; comparing positive vs negative Raoult deviations. Intext 7.1 lists concentration methods. Intext 7.2: Raoult's and Henry's laws. Intext 7.3: colligative properties and why π is best for biomolecules.
A solution is a homogeneous mixture. Concentration is expressed as M, m, N, mole fraction, or mass %. Henry's law governs gas solubility; Raoult's law links vapour pressure to mole fraction. Ideal solutions obey Raoult's law; real solutions show positive or negative deviations. Colligative properties — RLVP, ΔTb, ΔTf, π — depend on particle count; electrolyte dissociation and solute association are corrected using i. Together these concepts explain everyday phenomena from boiling point elevation in cooking to medical saline isotonicity and reverse-osmosis water purification.
The logical flow of Module 3 builds from defining what a solution is (7.1), through classifying types and gas solubility (7.2), to vapour pressure as the underlying property (7.3), Raoult's law as the quantitative link between composition and VP (7.4–7.6), and finally colligative properties as practical consequences of VP lowering (7.7–7.8). Mastering the derivations is less important than knowing which formula applies in which situation and whether the van't Hoff correction is needed — that decision tree is what NIOS terminal and board questions typically test.
Most exam-important points from this chapter:
M = n/V (T-dependent). m = n×1000/W_solvent (T-independent) — use m for colligative properties. N uses gram equivalents.
Henry: x ∝ p. Solubility decreases on heating. Valid only under moderate P, no reaction/association.
P_i = P_i°·X_i. Non-volatile solute: (P°−P)/P° = X_solute. Ideal = no ΔH/ΔV on mixing.
RLVP, ΔT_b=K_b m, ΔT_f=K_f m, π=CRT — all depend on particle count, not nature.
Electrolytes: i>1 (NaCl≈2). Association: i<1 (dimer). Always use i·K·m or i·CRT when particles change.
Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L7 — Solutions only. Use Model Answer for marking points; Explanation for concept clarity.
7 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A
PYQ1. Which of the following colligative properties can be used to determine molar mass of proteins with maximum precision? — (A) Depression in freezing point (B) Osmotic pressure (C) Relative lowering of vapour pressure (D) Elevation of boiling point {ZåZ{b{IV ‘
Model Answer
Answer: (B) Osmotic pressure
Explanation
π = CRT; osmotic pressure is preferred for macromolecules (large M, small m) — most precise among colligative methods.
Source paper: 313/TUS/105A · Q2 · 1 mark(s) · Chapter L7.
PYQ2. Write True (T) for correct statement and False (F) for incorrect statement (out of four attempt any two) : Benzene-chloroform mixture exhibits positive deviation from Raoult’s law. Boiling point is a colligative property. Vapour pressure of a liquid is the pressure exerted by the vapour of the liquid in any condition/situation. Two liquids are miscible when they dissolve in each other in all proportions. ghr
Model Answer
Answer using key concepts from L7 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.
Explanation
Cross-check with L7 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q19 · 2 mark(s) · L7.
PYQ3. The outer shells of two eggs of the same size have been removed. Using these, how will you prove that the membrane covering egg is a semipermeable membrane? 4% NaOH solution and 6% urea solution (weight/volume in both cases) are equimolar but not isotonic. Explain. g‘mZ
Model Answer
Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.
Explanation
Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q35 · 2 mark(s) · L7.
PYQ4. An aqueous solution is made by dissolving 10 g of glucose (C6H12O6) in 90 g of water at 300 K. What is the mole fraction of water in this solution? 300 K na 10 J«m‘ ½byH$mog (C6H12O6)
Model Answer
State the precise definition from the L7 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.
Explanation
Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.
How to write for NIOS: Use 50–80 words with equation + reason. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q39 · 3 mark(s) · L7.
PYQ5. The outer shells of two eggs of the same size have been removed. Using these, how will you prove that the membrane covering egg is a semipermeable membrane? 4% NaOH solution and 6% urea solution (weight/volume in both cases) are equimolar but not isotonic. Explain. g‘mZ
Model Answer
Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.
Explanation
Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q29 · 2 mark(s) · L7.
PYQ6. The outer shells of two eggs of the same size have been removed. Using these, how will you prove that the membrane covering egg is a semipermeable membrane? 4% NaOH solution and 6% urea solution (weight/volume in both cases) are equimolar but not isotonic. Explain. g‘mZ
Model Answer
Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.
Explanation
Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q32 · 2 mark(s) · L7.
PYQ7. The density of 2·0 M solution of acetic acid (molar mass = 60·0 g mol–1) is 1·02 g/mL. Calculate the molality of the solution. Egr{Q>H$ Aåb (‘moba Ðì¶‘mZ = 60·0 g mol–1) Ho$ 2·0 M {db¶Z
Model Answer
Answer using key concepts from L7 (definitions, equations, and one example where useful). Stay within the suggested word range for a 3-mark NIOS question.
Explanation
Cross-check with L7 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.
How to write for NIOS: Use 50–80 words with equation + reason. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q39 · 3 mark(s) · L7.
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.
Draw a simple solution sketch (solute/solvent). Calculate molarity if 0.50 mol NaCl is dissolved to make 0.25 L solution.
Final answer: 2.0 M
Molarity is moles of solute per litre of solution.
Working formulas: M = n_solute / V_solution (L). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Half a mole in a quarter litre → 2 moles per litre.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Temperature changes volume slightly, so M is T-dependent.
Linked to chapter notes (L7). Remember: M = n_solute / V_solution (L). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write M = n_solute / V_solution (L) before substituting. Keep three significant figures until the end when data allow.
2.0 mol of solute in 1.0 kg solvent: find molality.
Final answer: 2.0 molal
Molality uses solvent mass; independent of temperature.
Working formulas: m = n_solute / mass_solvent (kg). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Two moles per kilogram of solvent is 2 molal.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Prefer molality when T varies (colligative problems).
Linked to chapter notes (L7). Remember: m = n_solute / mass_solvent (kg). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write m = n_solute / mass_solvent (kg) before substituting. Keep three significant figures until the end when data allow.
A mixture has 1 mol A and 3 mol B. Mole fraction of A?
Final answer: x_A = 0.25
Mole fraction is dimensionless; sum of mole fractions = 1.
Working formulas: x_A = n_A / n_total. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
A is one of four moles total → 0.25.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Used in Raoult’s law for vapour pressure.
Linked to chapter notes (L7). Remember: x_A = n_A / n_total. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write x_A = n_A / n_total before substituting. Keep three significant figures until the end when data allow.
State Raoult’s law for a volatile component in an ideal solution.
Final answer: p_A = x_A p_A°
Ideal solutions obey Raoult’s law over all compositions.
Working formulas: p_A = x_A p_A°. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
More of A in liquid → more A vapour above, proportional to its fraction.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Deviations → non-ideal solutions (positive/negative).
Linked to chapter notes (L7). Remember: p_A = x_A p_A°. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write p_A = x_A p_A° before substituting. Keep three significant figures until the end when data allow.
Why does adding non-volatile solute raise boiling point of solvent?
Final answer: Lower p ⇒ higher b.p.; ΔT_b = K_b m
Colligative properties depend on number of solute particles, not identity (ideal dilute).
Working formulas: ΔT_b = K_b m. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Solute “blocks” solvent escape to vapour—so harder to boil.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
i factor (van’t Hoff) for electrolytes.
Linked to chapter notes (L7). Remember: ΔT_b = K_b m. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write ΔT_b = K_b m before substituting. Keep three significant figures until the end when data allow.
State the formula for osmotic pressure of a dilute solution.
Final answer: π = C R T
Osmosis: solvent flows through semipermeable membrane toward higher solute concentration.
Working formulas: π = C R T. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Pressure needed to stop pure solvent entering the solution is π.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Used to find molar masses of polymers/biomolecules.
Linked to chapter notes (L7). Remember: π = C R T. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write π = C R T before substituting. Keep three significant figures until the end when data allow.