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Chemistry — Class 12 — L7: Solutions

NIOS Code 313 · Module 3 · States of Matter

Notes extracted from NIOS Chemistry Course (313), Lesson 7 — Solutions (313_Chemistry_Eng_Lesson7.pdf). Content covers sections 7.1–7.8.
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Overview — Module 3: Solutions

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.

Section 1: Components and Concentration (7.1)

7.1 Components of a Solution

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.

7.1.1 Concentration Units

Properties like sweetness or colour depend on how much solute is present relative to solvent — the concentration. NIOS covers five main expressions:

  • Molarity (M): moles of solute per litre of solution. M = n/V. Example: 2.0 M H₂SO₄ has 2.0 mol per litre. Changes with temperature because solution volume expands or contracts.
  • Molality (m): moles of solute per kilogram of solvent. m = nB×1000/WA. Does not change with temperature — used in colligative property calculations.
  • Normality (N): gram equivalent weights per litre. Eq. wt = mol. wt / acidity (acid), basicity (base), or valency. N = strength (g/L) / eq. wt.
  • Mole fraction (x): xA = nA/(nA+nB); sum of all x = 1.
  • Mass percentage: grams of solute per 100 g solution (5% KMnO₄ = 5 g per 100 g solution).
M = n/V  |  m = nB×1000/WA
M = mol L⁻¹ (temperature-dependent) · m = mol kg⁻¹ solvent (temperature-independent) · Example 7.1: 32 g CH₃OH in 200 mL → 5 M

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.

Section 2: Types of Solutions and Henry's Law (7.2)

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:

  • Liquids in liquids: completely miscible (alcohol–water), partially miscible (water–phenol), or immiscible (water–benzene). Solubility generally increases with temperature.
  • Gases in liquids: O₂ dissolves enough for aquatic life; CO₂ and NH₃ are highly soluble. Depends on pressure, temperature, and nature of gas/solvent.
  • Solids in liquids: e.g. NaCl in water — extent of dissolution varies by substance.
Henry's Law: x = K·p
x = mole fraction of gas in solution · p = partial pressure · K = Henry's constant · Gas solubility ∝ pressure; decreases with rising temperature at constant P

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.

Section 3: Vapour Pressure and Raoult's Law (7.3–7.5)

7.3 Vapour 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).

Raoult's Law — Vapour Pressure vs Mole Fraction Mole fraction XA Vapour pressure PA° PB° P = PA + PB (ideal) PA = PA°·XA   PB = PB°·XB
Fig 7.2 — Ideal solution: total vapour pressure varies linearly between PA° and PB°.

7.4 Raoult's Law for Volatile Liquids

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.

7.5 Raoult's Law for Non-Volatile Solute

For aqueous sugar/salt solutions, only solvent vapour is present. Since XA < 1, vapour pressure drops below pure solvent. Derivation gives:

(PA° − P)/PA° = XB
Relative lowering of vapour pressure = mole fraction of non-volatile solute · MB = (WB·MA)/(WA·RLVP) · Example 7.5: RLVP 0.00715 → M ≈ 181

Section 4: Ideal and Non-Ideal Solutions (7.6)

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.

  • Positive deviation: A–B interactions weaker than A–A or B–B → total P higher than predicted (maximum at intermediate composition). Examples: ethanol–cyclohexane, water–propanol, acetone–CS₂.
  • Negative deviation: A–B interactions stronger → total P lower (minimum at intermediate composition). Examples: chloroform–acetone, water–H₂SO₄, phenol–aniline.

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.

Non-Ideal Deviations from Raoult's Law Positive (+) Negative (−)
Fig 7.3–7.4 — Positive: VP above ideal line. Negative: VP below ideal line.

Section 5: Colligative Properties (7.7)

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.

7.7.1 Relative Lowering of Vapour Pressure

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 → MB181 g mol⁻¹. VP lowering is the root cause of all other colligative effects — reduced solvent escaping tendency shifts every phase equilibrium.

7.7.2 Elevation of Boiling Point

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.

7.7.3 Depression in Freezing Point

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.

ΔTb = Kb·m  |  ΔTf = Kf·m
m = molality (mol kg⁻¹) · Example 7.6: glucose in water → ΔTb=0.018 K, ΔTf=0.66 K · Kf(H₂O)=1.86 K kg mol⁻¹

7.7.4 Osmosis and Osmotic Pressure

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.

Osmosis and Osmotic Pressure Pure solvent Solution SPM solvent flow Apply π to stop π = CRT — best method for molar mass of biomolecules (room temperature)
Fig 7.7 — Solvent flows into solution; osmotic pressure π prevents further rise.
π = CRT = (n/V)RT  |  M = (wRT)/(πV)
π in atm · R = 0.082 L atm K⁻¹ mol⁻¹ · Example 7.7: protein π → M = 61022 g mol⁻¹ · Reverse osmosis: P > π desalinates seawater

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.

Section 6: Abnormal Colligative Properties (7.8)

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).

i = observed colligative / normal colligative = Mnormal/Mobserved
Dissociation: i = 1 + x (KCl) · Association dimer: i = 1 − x/2 · Modified equations: ΔTb=iKbm, ΔTf=iKfm, πV=iCRT

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.

van't Hoff Factor — Particle Count Effect Dissociation NaCl → Na⁺+Cl⁻ i ≈ 2 No change Glucose, urea i = 1 Association 2 acid ⇌ dimer i < 1
Colligative property magnitude tracks effective number of solute particles via van't Hoff factor i.

Exam Connections and Chapter Summary

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.

MCQ Quiz — L7 Solutions

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Flashcards — L7

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Golden Rules — L7 Solutions

Most exam-important points from this chapter:

Concentration units

M = n/V (T-dependent). m = n×1000/W_solvent (T-independent) — use m for colligative properties. N uses gram equivalents.

Gas solubility

Henry: x ∝ p. Solubility decreases on heating. Valid only under moderate P, no reaction/association.

Raoult's law

P_i = P_i°·X_i. Non-volatile solute: (P°−P)/P° = X_solute. Ideal = no ΔH/ΔV on mixing.

Colligative four

RLVP, ΔT_b=K_b m, ΔT_f=K_f m, π=CRT — all depend on particle count, not nature.

van't Hoff i

Electrolytes: i>1 (NaCl≈2). Association: i<1 (dimer). Always use i·K·m or i·CRT when particles change.

M = n/V
m = nB×1000/WA
x = Kp (Henry)
PA = PA°·XA
(P°−P)/P° = XB
ΔTb = Kb·m
ΔTf = Kf·m
π = CRT
i = observed/normal

Section 1: Formulas with Derivations

All key equations from NIOS Chemistry 313, Module 3 — Solutions (sections 7.1–7.8). Unlock for full derivations, examples, and exam prep.

Molarity — M = n/V (mol L⁻¹)

Formula: M = n/V  |  n = mass/Molar mass

Where

• n = moles of solute
• V = volume of solution in litres
• M = molarity (mol L⁻¹)

When to use

Volumetric lab work, titrations — e.g. 32 g CH₃OH in 200 mL → 5 M (Example 7.1).

Memory aid: "Moles per Litre" — M = n/V.

❌ Using solvent volume instead of solution volume.
✓ M uses total solution volume.

Worked Examples

Basic

Q: Molarity of 0.5 mol in 250 mL?

M = 0.5/0.25

Answer: 2.0 M

Intermediate

Q: 32 g CH₃OH (M=32) in 200 mL solution?

n=1 mol; M=1/0.2

Answer: 5 M (Example 7.1)

Exam

Q: Why does molarity change on heating?

Volume expands with temperature; n fixed → M decreases

Answer: M is temperature-dependent

Molality — m = nB×1000/WA (mol kg⁻¹)

Formula: m = moles solute / kg solvent

Where

• WA = mass of solvent in grams
• Independent of temperature — use for colligative properties

When to use

ΔTb, ΔTf, π calculations; Example 7.2: 50% H₂SO₄ → m ≈ 6.8 m.

Memory aid: "Molal = moles per kilogram solvent" — small m, big difference from M.

❌ Using molarity in colligative formulas.
✓ Always use molality (m) for ΔTb, ΔTf, π.

Worked Examples

Intermediate

Q: Molality if 10 g NaOH in 500 g water?

n = 10/40 = 0.25; m = 0.25/0.5

Answer: 0.5 m

Exam

Q: 36 g water + 46 g ethanol — mole fractions?

nw=2, ne=1; xw=2/3, xe=1/3

Answer: 0.67 and 0.33 (Example 7.3)

Henry's Law — x = K·p

Formula: Mole fraction of dissolved gas ∝ partial pressure above liquid

Valid when

Moderate pressure · Not too low temperature · No association/dissociation/reaction with solvent

Real-world application

Carbonated drinks (CO₂ under pressure); scuba diving (N₂ dissolution); fish need dissolved O₂.

Intermediate

Q: Why does warm soda go flat faster?

Gas solubility decreases with temperature at constant P

Answer: CO₂ solubility drops: 0.88→0.53 (20°C→40°C)

Raoult's Law — Volatile Liquids

PA = PA°·XA  |  PB = PB°·XB

Total: P = PA°·XA + PB°·XB

Ideal solution

Obeys Raoult's law at all concentrations; ΔHmix=0, ΔVmix=0

Non-volatile solute

(P°−P)/P° = XB — relative lowering of vapour pressure

Derivation (non-volatile)

Only solvent evaporates. P = PA°·XA and XA + XB = 1, so P/P° = XA = 1 − XB. Hence RLVP = XB.

Exam

Q: RLVP = 0.00715 — find molar mass of solute (Example 7.5)

MB = (WB·MA)/(WA·RLVP)

Answer: M ≈ 181 g mol⁻¹

Colligative Properties

ΔTb = Kb·m — elevation in boiling point
ΔTf = Kf·m — depression in freezing point
π = CRT = (n/V)RT — osmotic pressure (R = 0.0821 L atm K⁻¹ mol⁻¹)

Key principle

Depend only on number of solute particles, not their nature. More particles → greater effect.

Memory aid: "Colligative = collective count" — count particles, not identity.

❌ Forgetting van't Hoff factor i for electrolytes.
✓ Use i·m or i·C in all colligative equations when solute dissociates.

Worked Examples

Basic

Q: ΔTb for 1 m glucose in water (Kb=0.52)?

ΔTb = 0.52 × 1

Answer: 0.52 K

Intermediate

Q: Freezing point of 0.5 m NaCl (Kf=1.86, i≈2)?

ΔTf = i·Kf·m = 2×1.86×0.5

Answer: ΔTf = 1.86 K below 0°C

Exam

Q: Molar mass from π = 2.5 atm at 300 K, C = 0.01 M?

π = CRT → verify; or use πV = nRT with mass unknown

Answer: Set up π = (w/M)×(1/V)×RT, solve for M

van't Hoff Factor — i

i = observed colligative / normal colligative
i = normal M / observed M

NaCl dissociation: i = 1 + α (α = degree of dissociation)
Association (benzoic acid in benzene): i < 1

Modified equations

ΔTb = iKbm · ΔTf = iKfm · π = iCRT · RLVP uses effective mole fraction

Exam

Q: KCl — i = 1.92. Degree of dissociation?

KCl → K⁺ + Cl⁻; i = 1 + α; α = 0.92

Answer: 92% dissociated

Section 2: Detailed Definitions

DEFINITION: Solution

Meaning: Homogeneous mixture of two or more substances.
Components: Solvent (major, same phase as solution) + solute (dissolved)
Example: Sugar in water — water is solvent
Common confusion: In gas–gas (air), major component is solvent by convention.

DEFINITION: Mole Fraction (x)

Formula: xA = nA/(nA+nB)
Property: Sum of all mole fractions = 1; no units
Use: Raoult's law, vapour pressure calculations
Common confusion: Mole fraction ≠ mass percent.

DEFINITION: Colligative Property

Meaning: Property depending on number of solute particles, not identity.
Four types: RLVP, ΔTb, ΔTf, π
Example: 1 m glucose and 0.5 m NaCl can give same ΔTb if i=2 for NaCl
Connects to: van't Hoff factor for electrolytes

DEFINITION: Ideal Solution

Meaning: Obeys Raoult's law at all concentrations and temperatures.
Conditions: ΔHmix = 0, ΔVmix = 0
Example: Benzene–toluene (similar intermolecular forces)
Non-ideal: Positive deviation (ethanol–cyclohexane); negative (CHCl₃–acetone)

DEFINITION: Osmosis

Meaning: Solvent flows through semipermeable membrane from lower to higher concentration.
π: Pressure needed to stop osmosis
Applications: IV saline (isotonic), reverse osmosis desalination
Common confusion: Osmosis = solvent movement, not solute.

Section 3: Diagrams & Visuals

Solutions Formula Map — L7 M = n/V m = n/kg solv x = n/Σn Vapour pressure: P_A = P°·X_A (Raoult) RLVP = X_B (non-vol) Colligative (use molality + i): ΔT_b = iK_b m · ΔT_f = iK_f m · π = iCRT

Concentration units → vapour pressure → colligative properties (all linked via particle count)

COLLIGATIVE PROPERTY CHAIN ═══════════════════════════════════════ Add non-volatile solute │ ▼ RLVP = X_B (fewer solvent molecules at surface) │ ┌────┴────┬────────┐ ▼ ▼ ▼ ΔT_b↑ ΔT_f↓ π (osmosis) │ │ │ └──── i factor for electrolytes ────┘ Molar mass from any colligative measurement ═══════════════════════════════════════

Section 5: Comprehensive Q&A (12 Questions)

Q1: Molarity vs molality — when to use which?

Molarity (M) for lab volumetric work — titrations, solution prep by volume. Molality (m) for colligative properties and when temperature varies — m uses mass of solvent, unaffected by expansion.

Q2: State Raoult's law for ideal binary liquid solution.

PA = PA°·XA and PB = PB°·XB. Total P = PA°·XA + PB°·XB. Partial pressure proportional to mole fraction.

Q3: Why does adding salt lower the freezing point of water?

Solute particles disrupt ice crystal formation. ΔTf = Kf·m — more particles → lower freezing point. Road salt exploits this in winter.

Q4: What is relative lowering of vapour pressure?

(P°−P)/P° = XB for non-volatile solute. Fraction by which vapour pressure drops equals mole fraction of solute. Independent of nature of solute (colligative).

Q5: Positive vs negative deviation from Raoult's law?

Positive: A–B interactions weaker than A–A, B–B (ethanol–cyclohexane) — higher vapour pressure than ideal. Negative: A–B stronger (CHCl₃–acetone) — lower vapour pressure, may form maximum boiling azeotrope.

Q6: Henry's law and scuba diving — connection?

N₂ dissolves in blood under high pressure (x = Kp). Rapid ascent → pressure drops → N₂ comes out of solution (bends). Slow ascent allows gradual degassing.

Q7: How to calculate molar mass from osmotic pressure?

πV = nRT = (w/M)RT. Measure π, know w, V, T → solve for M. Very sensitive for large macromolecules (polymers, proteins).

Q8: What is isotonic solution?

Same osmotic pressure as body fluids (blood ~0.9% NaCl). IV drips must be isotonic to prevent cell bursting (hypotonic) or shrinking (hypertonic).

Q9: Why is observed molar mass of KCl less than 74.5?

KCl dissociates: K⁺ + Cl⁻. Colligative effect doubled → i ≈ 2 → apparent M ≈ 37. Abnormal colligative properties reveal dissociation.

Q10: Normality of 0.4 g NaOH in 100 mL?

Eq. wt NaOH = 40. Strength = 4 g/L. N = 4/40 = 0.1 N (Example 7.4).

Q11: Reverse osmosis — how does it work?

Apply external pressure greater than osmotic pressure π on concentrated side. Forces solvent through semipermeable membrane from high to low concentration — desalination.

Q12: Exam — 1 m glucose vs 0.5 m NaCl: which has higher ΔTb?

Glucose: i=1, effective particles = 1 m. NaCl: i≈2, effective = 1 m. Both give same ΔTb ≈ Kb×1.

Section 6: Tips, Tricks & Exam Hacks

Memory Aids

  • M vs m: "Molarity = Volume; Molality = kilogram"
  • Colligative: "Count particles, not personality"
  • Raoult volatile: "P proportional to X"
  • RLVP: "Drop equals X of solute"
  • i for NaCl: "i ≈ 2 — two ions"

Exam Tips

  • Colligative problems: always check if electrolyte → use i
  • Molar mass from RLVP: MB = (WB·MA)/(WA·RLVP)
  • π in atm: R = 0.0821 L atm K⁻¹ mol⁻¹
  • Convert mL to L and g to kg carefully
  • Ideal solution: straight line on P vs X graph

Section 7: Connections & Relationships

This chapter builds on: L1 mole concept (n = m/M), L4 intermolecular forces (H-bonding, dipole interactions in non-ideal solutions).
This chapter leads to: L12 ionic equilibrium (activity, ionic strength), L13 electrochemistry (conductivity of solutions), biochemistry (isotonic fluids).
Related formulas: Molarity from L1; Born–Haber dissolution steps; van't Hoff links to equilibrium dissociation.

CONCENTRATION UNIT CHOICES Lab titration (fixed volume) ──► Molarity M Colligative / temperature change ──► Molality m Vapour pressure / distillation ──► Mole fraction x Acid-base redox equivalents ──► Normality N Mass-based recipes ──► Mass %

Section 8: Complete Quick Reference

Formulas at a glance:

• M = n/V — molarity (mol L⁻¹)

• m = n×1000/WA — molality (mol kg⁻¹)

• xA = nA/Σn — mole fraction

• N = strength/eq. wt — normality

• x = Kp — Henry's law

• PA = PA°·XA — Raoult (volatile)

• (P°−P)/P° = XB — RLVP (non-volatile)

• ΔTb = iKbm · ΔTf = iKfm · π = iCRT

• i = observed/normal colligative

Key terms: Ideal solution · Azeotrope · Osmosis · Isotonic · Colligative · van't Hoff factor

Decision tree: Concentration? → pick M/m/x/N. Vapour pressure? → Raoult. Non-volatile solute? → RLVP. Boiling/freezing/osmosis? → colligative with m and i.

Remember: ✓ Colligative uses m not M ✓ Electrolytes need i ✓ M changes with T ✓ Osmosis = solvent flow ✓ Ideal: ΔHmix=0

PYQ — Previous Year Questions

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.

L7 — Solutions

7 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

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 ‘

1 mark · Q2 · 313/TUS/105A

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

2 marks · Q19 · 313/TUS/105A

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.

Section B — Short / Long answer (from papers)

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

2 marks · Q35 · 313/MAY/205A

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)

3 marks · Q39 · 313/MAY/205A

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

2 marks · Q29 · 313/MAY/205B

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

2 marks · Q32 · 313/MAY/205C

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

3 marks · Q39 · 313/MAY/205C

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.

Problem Solving — L7 Solutions

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.

Question 1 of 6Molarity

Draw a simple solution sketch (solute/solvent). Calculate molarity if 0.50 mol NaCl is dissolved to make 0.25 L solution.

M = n_solute / V_solution (L)

Pencil sketch (labelled)

Solution components solute solvent M = n_solute / V_solution (L)
Pencil sketch: solute particles in solvent

Solution — step by step with formulas

  1. M = 0.50 / 0.25 = 2.0 mol L⁻¹.

Final answer: 2.0 M

Formulas used in this problem

M = n_solute / V_solution (L)

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Molarity

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.

Exam tip

Write M = n_solute / V_solution (L) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Molality

2.0 mol of solute in 1.0 kg solvent: find molality.

m = n_solute / mass_solvent (kg)

Solution — step by step with formulas

  1. m = 2.0 mol kg⁻¹.

Final answer: 2.0 molal

Formulas used in this problem

m = n_solute / mass_solvent (kg)

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Molality

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.

Exam tip

Write m = n_solute / mass_solvent (kg) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Mole fraction

A mixture has 1 mol A and 3 mol B. Mole fraction of A?

x_A = n_A / n_total

Solution — step by step with formulas

  1. x_A = 1/4 = 0.25.

Final answer: x_A = 0.25

Formulas used in this problem

x_A = n_A / n_total

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Mole fraction

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.

Exam tip

Write x_A = n_A / n_total before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Raoult

State Raoult’s law for a volatile component in an ideal solution.

p_A = x_A p_A°

Solution — step by step with formulas

  1. Partial vapour pressure = mole fraction × pure vapour pressure.

Final answer: p_A = x_A p_A°

Formulas used in this problem

p_A = x_A p_A°

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Raoult’s law

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.

Exam tip

Write p_A = x_A p_A° before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Colligative

Why does adding non-volatile solute raise boiling point of solvent?

ΔT_b = K_b m

Solution — step by step with formulas

  1. Vapour pressure lowers; higher T needed to reach atmospheric pressure ⇒ ΔT_b = K_b m.

Final answer: Lower p ⇒ higher b.p.; ΔT_b = K_b m

Formulas used in this problem

ΔT_b = K_b m

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Elevation of boiling point

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.

Exam tip

Write ΔT_b = K_b m before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Osmosis

State the formula for osmotic pressure of a dilute solution.

π = C R T

Solution — step by step with formulas

  1. π = CRT (C molarity, T absolute).

Final answer: π = C R T

Formulas used in this problem

π = C R T

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Osmotic pressure

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.

Exam tip

Write π = C R T before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.