Class 12 Chemistry · Chapter 1
Short answer:
Ye chapter NCERT Class 12 Chemistry Part I ka Chapter 1 hai. Isme 20 exercise-style questions cover kiye gaye hain jo concentration terms (mass %, mole fraction, molarity, molality, ppm), solubility aur Henry's law, Raoult's law (ideal aur non-ideal solutions), colligative properties (relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, osmotic pressure) aur van't Hoff factor / abnormal molar mass jaise sub-topics ko span karte hain.
Solutions chapter physical chemistry ka backbone hai jo har numerical-heavy paper me directly aata hai. Board exam me concentration aur colligative property numericals guaranteed hain, aur NEET/JEE me van't Hoff factor + osmotic pressure ke questions har saal repeat hote hain — isliye formula aur unit dono pakka karna zaroori hai.
Chapter 1 Summary — 5 Minute Revision
1. Types of Solutions
Solution ek homogeneous mixture hai do ya zyada components ka. Jo component zyada quantity me ho wo solvent hai, baaki solute. Physical state ke basis par 9 types possible hain (gas-in-gas, liquid-in-gas, solid-in-gas, gas-in-liquid, liquid-in-liquid, solid-in-liquid, gas-in-solid, liquid-in-solid, solid-in-solid) — hum mainly solid/liquid/gas solute in liquid solvent wale binary solutions padhte hain.
2. Concentration Terms
- Mass percentage (w/w): (mass of component / total mass of solution) × 100
- Volume percentage (v/v): liquid-liquid solutions ke liye
- ppm (parts per million): jab solute bahut kam ho (traces), mass fraction × 106
- Mole fraction (x): xA = nA/(nA+nB) — temperature-independent, isliye vapour pressure jaisi calculations me favourite hai
- Molarity (M): nsolute/V(litres) — temperature ke saath change hoti hai kyunki volume expand/contract karta hai
- Molality (m): nsolute/mass of solvent (kg) — temperature-independent kyunki ye mass par based hai, volume par nahi. Isliye temperature-sensitive experiments (jaise freezing/boiling point studies) me molality zyada reliable hai.
3. Solubility aur Henry's Law
Solid-in-liquid solubility temperature ke saath generally badhti hai (agar dissolution endothermic hai). Gas-in-liquid solubility temperature badhne par ghatt-ti hai — isi wajah se garam pani me dissolved oxygen kam hoti hai aur aquatic life disturb hoti hai, aur cold drinks kholne par CO2 zyada fizz karti hai.
Henry's Law: gas ki partial pressure uske mole fraction ke directly proportional hoti hai — p = KH x. Bada KH matlab kam solubility. Applications: soda bottles (high pressure se CO2 zyada dissolve), deep-sea diving (N2 narcosis/bends se bachne ke liye He-O2 mixture), high-altitude par kam O2 partial pressure se anoxia.
4. Vapour Pressure aur Raoult's Law
Raoult's Law (volatile-volatile solutions): har component ki partial vapour pressure uske mole fraction × pure component ki vapour pressure ke barabar hoti hai — p1 = p1° x1, aur total pressure ptotal = p1°x1 + p2°x2.
Non-volatile solute wale solution ke liye: Raoult's law ka special case — relative lowering of vapour pressure solute ke mole fraction ke barabar hoti hai: (p1° − p1)/p1° = x2.
5. Ideal aur Non-ideal Solutions
| Property | Ideal Solution | Non-ideal (Positive deviation) | Non-ideal (Negative deviation) |
|---|---|---|---|
| Raoult's law | obey karta hai poori range me | obey nahi karta | obey nahi karta |
| A-B interaction | A-A ≈ B-B ≈ A-B | A-B < A-A, B-B (weak) | A-B > A-A, B-B (strong) |
| ΔHmix | 0 | positive (endothermic) | negative (exothermic) |
| ΔVmix | 0 | positive | negative |
| Example | benzene + toluene | ethanol + acetone | chloroform + acetone |
↔ Table ko side me swipe karein
Azeotropes — mixtures jo constant boiling point par same composition me distill hote hain, isliye fractional distillation se separate nahi ho sakte. Minimum boiling azeotrope (positive deviation, e.g. ethanol-water 95.6%) aur maximum boiling azeotrope (negative deviation, e.g. HCl-water).
6. Colligative Properties
Ye properties sirf solute particles ki number par depend karti hain, unki nature par nahi:
- Relative lowering of vapour pressure: (p° − p)/p° = x2
- Elevation in boiling point: ΔTb = Kb m
- Depression in freezing point: ΔTf = Kf m
- Osmotic pressure: π = CRT, jahan C = molarity. Osmosis = solvent molecules ka semipermeable membrane ke through kam concentration se zyada concentration ki taraf movement. Reverse osmosis (RO) tab hota hai jab solution par osmotic pressure se zyada external pressure lagayi jaaye — isse solvent ulta direction me move karta hai (desalination me use hota hai).
Isotonic solutions wo hote hain jinka osmotic pressure equal ho (koi net osmosis nahi). Hypertonic solution me cell shrink (plasmolysis) hoti hai, hypotonic me cell swell/burst (haemolysis) hoti hai.
7. van't Hoff Factor aur Abnormal Molar Mass
Electrolytes dissociate hote hain (jaise NaCl → Na+ + Cl−) to expected se zyada particles banate hain, aur kuch solutes (jaise benzoic acid benzene me) associate/dimerise hote hain to expected se kam particles banate hain. Isse observed colligative property calculated se different aa jaati hai — is deviation ko van't Hoff factor i capture karta hai:
- i = observed colligative property / calculated colligative property (assuming no dissociation)
- i = normal molar mass / observed (abnormal) molar mass
- Dissociation ke liye: i = 1 + (n−1)α, jahan n = ions ki number aur α = degree of dissociation
- Association ke liye: i = 1 + (1/n − 1)α, jahan n = association ka degree (e.g. dimer ke liye n=2)
- i > 1 → dissociation, i < 1 → association, i = 1 → non-electrolyte, no association
Poore Class 12 Chemistry ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q20)
Q1. 22 g benzene (C6H6) ko 122 g carbon tetrachloride (CCl4) me dissolve kiya gaya hai. Dono ki mass percentage calculate karo.
Total mass of solution = 22 + 122 = 144 g
Mass % of C6H6 = (22/144) × 100 = 15.28%
Mass % of CCl4 = (122/144) × 100 = 84.72%
Q2. Ek solution CCl4 me 30% (by mass) benzene ka hai. Benzene ka mole fraction nikalo. (Mbenzene = 78 g/mol, MCCl4 = 154 g/mol)
Maan lo 100 g solution: 30 g benzene + 70 g CCl4
n(benzene) = 30/78 = 0.3846 mol
n(CCl4) = 70/154 = 0.4545 mol
x(benzene) = 0.3846/(0.3846+0.4545) = 0.3846/0.8391 = 0.458
x(CCl4) = 1 − 0.458 = 0.542
Q3. 30 g ethanol (C2H5OH, M = 46 g/mol) ko pani me dissolve karke 500 mL solution banaya gaya. Molarity nikalo.
Given: mass = 30 g, M = 46 g/mol, V = 500 mL = 0.500 L
n = 30/46 = 0.6522 mol
Molarity = n/V = 0.6522/0.500 = 1.30 mol/L
Q4. 2.5 g ethanoic acid (CH3COOH, M = 60 g/mol) ko 75 g benzene me dissolve kiya gaya. Molality nikalo.
n(CH3COOH) = 2.5/60 = 0.04167 mol
mass of solvent = 75 g = 0.075 kg
molality = 0.04167/0.075 = 0.556 mol/kg
Q5. 0.10 g oxygen gas ko 1 kg pani me dissolve kiya gaya. Concentration ko ppm me express karo.
Total mass of solution ≈ 1000 + 0.10 = 1000.1 g
ppm = (mass of solute/mass of solution) × 106 = (0.10/1000.1) × 106 ≈ 100 ppm
Q6. 298 K par methane ki solubility benzene me measure ki gayi: mole fraction x = 3.5 × 10−3 jab partial pressure p = 12.8 bar hai. Henry's law constant KH nikalo, aur phir 1 atm (1.013 bar) pressure par methane ka mole fraction nikalo.
Henry's law: p = KHx
KH = p/x = 12.8/(3.5×10−3) = 3657 bar
At p = 1.013 bar: x = p/KH = 1.013/3657 = 2.77 × 10−4
Q7. Temperature badhne par gases ki liquid me solubility kyun kam ho jaati hai? Iska ek real-life example do.
Gas ka liquid me dissolve hona ek exothermic process hai (gas molecules solvent me settle hote waqt energy release karte hain). Le Chatelier's principle ke according, temperature badhane se equilibrium reverse direction (gas nikalne) ki taraf shift hota hai — isliye solubility ghat jaati hai.
Example: garam pani me dissolved oxygen ki matra thandi pani se kam hoti hai, jo garmi ke mausam me machhliyon aur aquatic life ke liye stressful hoti hai.
Q8. Raoult's law ko volatile-volatile binary solution ke liye state karo. Ideal solution kise kehte hain?
Raoult's law: kisi bhi component ki solution me partial vapour pressure, uske mole fraction aur pure state me uski vapour pressure ke product ke barabar hoti hai: p1 = p1°x1, p2 = p2°x2, aur ptotal = p1°x1 + p2°x2.
Ideal solution wo hota hai jo poori concentration range me Raoult's law obey kare, jismein A-A, B-B aur A-B interactions almost equal hon, ΔHmix = 0 aur ΔVmix = 0 ho (e.g. benzene + toluene, n-hexane + n-heptane).
Q9. 100 g liquid A (M = 140 g/mol) ko 1000 g solvent B (M = 180 g/mol) me dissolve kiya gaya. Pure B ki vapour pressure 500 torr hai. Solution ki total vapour pressure 475 torr hai. Pure A ki vapour pressure nikalo.
n(A) = 100/140 = 0.714 mol; n(B) = 1000/180 = 5.556 mol
x(A) = 0.714/6.270 = 0.1139; x(B) = 0.8861
p(B) = p°(B) × x(B) = 500 × 0.8861 = 443.05 torr
p(A) = 475 − 443.05 = 31.95 torr
p°(A) = p(A)/x(A) = 31.95/0.1139 = ≈ 280 torr
Q10. 298 K par pure pani ki vapour pressure 23.8 mm Hg hai. 50 g urea (M = 60 g/mol) ko 850 g pani me dissolve kiya gaya. Solution ki vapour pressure nikalo.
n(urea) = 50/60 = 0.8333 mol; n(water) = 850/18 = 47.22 mol
x(urea) = 0.8333/(0.8333+47.22) = 0.8333/48.06 = 0.01734
(p° − p)/p° = x(urea) ⟹ Δp = 23.8 × 0.01734 = 0.4127 mm Hg
p(solution) = 23.8 − 0.4127 = 23.39 mm Hg
Q11. Raoult's law se positive aur negative deviations ka matlab samjhao, ek-ek example ke saath.
Positive deviation: jab solution ki vapour pressure Raoult's law ke prediction se zyada hoti hai — A-B interactions A-A/B-B se weaker hote hain, molecules ek doosre ko chhodne me easy feel karte hain. ΔHmix positive (endothermic), ΔVmix positive. Example: ethanol + acetone.
Negative deviation: jab vapour pressure prediction se kam hoti hai — A-B interactions A-A/B-B se stronger hote hain (jaise H-bonding), molecules escape karne me resist karte hain. ΔHmix negative (exothermic), ΔVmix negative. Example: chloroform + acetone.
Q12. Azeotrope kya hota hai? Minimum aur maximum boiling azeotropes ka ek-ek example do.
Azeotrope binary liquid mixtures hote hain jo constant boiling point par boil hote hain aur liquid ki tarah hi composition me vapour banate hain — isliye fractional distillation se inke components separate nahi ho sakte.
- Minimum boiling azeotrope (positive deviation wale solutions se): ethanol-water (~95.6% ethanol)
- Maximum boiling azeotrope (negative deviation wale solutions se): HCl-water (~20.2% HCl)
Q13. 25°C par pure benzene ki vapour pressure 639.7 mm Hg hai. 39.0 g benzene me 2.175 g ek non-volatile solute dissolve karne par solution ki vapour pressure 631.9 mm Hg ho jaati hai. Solute ka molar mass nikalo.
Δp = 639.7 − 631.9 = 7.8 mm Hg
Δp/p° = 7.8/639.7 = 0.012193
n(benzene) = 39.0/78 = 0.500 mol
dilute solution: x2 ≈ n2/n1 ⟹ n2 = 0.012193 × 0.500 = 0.006097 mol
M2 = 2.175/0.006097 = ≈ 357 g/mol
Q14. 2 g Na2SO4 (M = 142 g/mol) ko 500 g pani me dissolve kiya gaya, poora dissociation maan kar (Na2SO4 → 2Na+ + SO42−, i = 3). Solution ka boiling point nikalo. (Kb for water = 0.52 K kg/mol)
n = 2/142 = 0.01408 mol
molality = 0.01408/0.500 = 0.02817 mol/kg
ΔTb = i × Kb × m = 3 × 0.52 × 0.02817 = 0.0440 K
Boiling point = 100 + 0.044 = 100.044°C
Q15. 1.00 g ek non-electrolyte solute ko 50 g benzene me dissolve karne se benzene ka freezing point 0.40 K se kam ho jaata hai. (Kf for benzene = 5.12 K kg/mol). Solute ka molar mass nikalo.
m = ΔTf/Kf = 0.40/5.12 = 0.07813 mol/kg
n(solute) = 0.07813 × 0.050 kg = 0.003906 mol
Molar mass = 1.00/0.003906 = ≈ 256 g/mol
Q16. 500 g pani ka freezing point 0.3 K se depress karne ke liye kitna KCl (M = 74.5 g/mol) chahiye hoga? (Kf for water = 1.86 K kg/mol; KCl fully dissociate hota hai, i = 2)
ΔTf = i × Kf × m ⟹ m = 0.3/(2 × 1.86) = 0.08065 mol/kg
n(KCl) = 0.08065 × 0.500 = 0.04032 mol
mass = 0.04032 × 74.5 = ≈ 3.0 g
Q17. 200 cm3 aqueous solution me 1.26 g ek non-volatile solute hai, jiska osmotic pressure 300 K par 0.259 atm hai. Solute ka molar mass nikalo. (R = 0.0821 L atm mol−1 K−1)
π = CRT ⟹ C = π/(RT) = 0.259/(0.0821 × 300) = 0.010516 mol/L
n = C × V = 0.010516 × 0.200 = 0.0021033 mol
Molar mass = 1.26/0.0021033 = ≈ 599 g/mol
Q18. 0.6 mL ethanoic acid (density = 1.06 g/mL) ko 1 L pani me dissolve kiya gaya. Isse freezing point 0.0205°C depress hota hai. Degree of dissociation nikalo. (Kf for water = 1.86 K kg/mol, M(CH3COOH) = 60 g/mol)
mass = 0.6 × 1.06 = 0.636 g; n = 0.636/60 = 0.0106 mol
molality ≈ 0.0106 mol/kg (1 L water ≈ 1 kg)
ΔTf(calculated, no dissociation) = Kf × m = 1.86 × 0.0106 = 0.01972 K
i = ΔTf(observed)/ΔTf(calculated) = 0.0205/0.01972 = 1.040
CH3COOH ⇌ CH3COO− + H+ (n=2 particles), isliye i = 1 + α
α = i − 1 = 0.040 ⟹ degree of dissociation ≈ 4.0%
Q19. 19.5 g CH2FCOOH (M = 78 g/mol) ko 500 g pani me dissolve karne par freezing point depression 1.0 K hoti hai. Acid ka degree of dissociation nikalo. (Kf for water = 1.86 K kg/mol)
n = 19.5/78 = 0.250 mol
molality = 0.250/0.500 = 0.500 mol/kg
ΔTf(calculated) = Kf × m = 1.86 × 0.500 = 0.930 K
i = ΔTf(observed)/ΔTf(calculated) = 1.0/0.930 = 1.0753
α = i − 1 = 0.0753 ⟹ ≈ 7.53%
Q20. Abnormal molar mass kya hoti hai? Ek example do jaha van't Hoff factor 1 se kam hota hai aur samjhao kyun.
Jab kisi solute ka experimentally observed molar mass, uske normal (formula-based) molar mass se different aata hai, to use abnormal molar mass kehte hain. Ye tab hota hai jab solute solution me dissociate ya associate karta hai — dissociation se particles ki number badhti hai (i > 1), aur association se particles ki number ghatti hai (i < 1).
Example: benzoic acid (C6H5COOH) benzene jaise non-polar solvent me hydrogen-bonding ki wajah se dimerise ho jaata hai (do molecules mil kar ek dimer banate hain). Isse effective particle count aadha ho jaata hai, isliye observed molar mass normal se roughly double aati hai aur van't Hoff factor i ≈ 0.5 (1 se kam) milta hai.
Important Equations — Ek Nazar Me
| Quantity | Formula | Notes |
|---|---|---|
| Mass percentage | (mass of component / total mass of solution) × 100 | temperature-independent |
| ppm | (mass of component / total mass of solution) × 106 | trace amounts ke liye |
| Mole fraction | xA = nA/(nA+nB) | sum of all x = 1; temperature-independent |
| Molarity (M) | M = nsolute/Vsolution(L) | temperature-dependent (volume badalta hai) |
| Molality (m) | m = nsolute/mass of solvent(kg) | temperature-independent |
| Henry's Law | p = KH x | bada KH = kam solubility |
| Raoult's Law (volatile-volatile) | p = p1°x1 + p2°x2 | ideal solution ke liye poori range me valid |
| Relative lowering of vapour pressure | (p1° − p1)/p1° = x2 | non-volatile solute case |
| Elevation in boiling point | ΔTb = i Kb m | Kb = molal elevation constant (K kg/mol) |
| Depression in freezing point | ΔTf = i Kf m | Kf = molal depression constant (K kg/mol) |
| Osmotic pressure | π = C R T | C = molarity; R = 0.0821 L atm mol−1 K−1 (ya 8.314 J mol−1 K−1 SI units me) |
| van't Hoff factor | i = observed colligative property / calculated colligative property = normal molar mass / observed molar mass | i > 1 dissociation, i < 1 association, i = 1 non-electrolyte |
| i for dissociation | i = 1 + (n−1)α | n = ions produced per formula unit |
| i for association | i = 1 + (1/n − 1)α | n = molecules per associated unit (e.g. dimer n=2) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Molarity aur molality ko confuse karna. Molarity volume (L) par based hai jo temperature ke saath expand/contract hoti hai; molality mass (kg) par based hai jo constant rehta hai. Colligative property numericals me hamesha molality use karo jab tak molarity specifically na di ho.
- Electrolyte solutions me van't Hoff factor bhool jaana. NaCl, Na2SO4, KCl jaise ionic compounds ke liye ΔTb, ΔTf aur π ke formula me i lagana zaroori hai (poora dissociation maan kar i = number of ions), warna answer galat aayega.
- Kb aur Kf ki units galat use karna. Ye K kg mol−1 (ya °C kg mol−1) me hoti hain, molality (mol/kg) ke saath multiply hoti hain — molarity ke saath nahi.
- Mole fraction calculate karte waqt moles nikalna bhool jaana. Directly masses ka ratio le lena galat hai — pehle har component ke moles (mass/molar mass) nikalo, tabhi mole fraction sahi aayega.
- Relative lowering of vapour pressure me sign/direction ki galti. (p° − p)/p° hamesha positive hota hai kyunki solute add karne se vapour pressure kam hoti hai (non-volatile solute case) — p ko p° se bada likh dena common mistake hai.
- Osmotic pressure formula me R ki value ki unit mismatch. Agar volume litres me aur pressure atm me hai to R = 0.0821 L atm mol−1 K−1 use karo; SI units (Pa, m3) me R = 8.314 J mol−1 K−1. Dono mix mat karo.
Board-Style Important Questions
- 1 mark: Molarity aur molality me se kaunsi quantity temperature-independent hai, aur kyun?
- 1 mark: Henry's law ka statement likho.
- 2 marks: Ideal aur non-ideal solution me antar batao, ek-ek example ke saath.
- 2 marks: van't Hoff factor kya darshata hai? Isse dissociation aur association ke case me kaise interpret karte hain?
- 3 marks: 100 g glucose (M = 180 g/mol) ko 1000 g pani me dissolve kiya gaya. Solution ki molality aur mole fraction of glucose calculate karo.
- 5 marks: Colligative properties kya hoti hain? In char colligative properties ke naam aur formula likho, aur samjhao ki inse molar mass kaise determine kiya jaata hai.
Aksar Poochhe Jaane Wale Sawaal
Molarity aur molality me kya farak hai?
Molarity (M) solute ke moles ko solution ke total volume (litres) se divide karke nikalte hain, jabki molality (m) solute ke moles ko solvent ke mass (kg) se divide karke. Molarity temperature ke saath change hoti hai (volume thermal expansion se badalta hai), lekin molality temperature-independent hoti hai kyunki mass fix rehta hai.
Henry's law kis cheez ke liye use hoti hai?
Henry's law gas ki liquid me solubility batati hai — gas ki partial pressure uske mole fraction ke proportional hoti hai (p = KHx). Isse deep-sea diving me nitrogen narcosis samajhna, soda bottles me CO2 ka behaviour, aur high altitude par kam oxygen availability jaise real applications explain hote hain.
Raoult's law aur ideal solutions ka kya connection hai?
Ideal solution wo hai jo poori concentration range me Raoult's law follow kare, matlab uski vapour pressure Raoult's law ke prediction ke exactly barabar ho. Aisa tab hota hai jab dono components ke beech intermolecular forces almost equal hon (A-A ≈ B-B ≈ A-B).
Colligative properties number of particles par kyun depend karti hain, unki nature par nahi?
Colligative properties (vapour pressure lowering, boiling point elevation, freezing point depression, osmotic pressure) solvent ke molecules ke escape/freeze/flow karne ki tendency par depend karti hain, jo solute particles ki sirf count se affect hoti hai — chahe wo particle glucose ho ya NaCl ka ion, effect similar hota hai bas count matter karta hai.
Van't Hoff factor 1 se zyada ya kam kyun hota hai?
i > 1 tab hota hai jab solute dissociate hota hai (jaise NaCl → Na+ + Cl−), kyunki particles ki actual number badh jaati hai. i < 1 tab hota hai jab solute associate/dimerise hota hai (jaise benzoic acid benzene me), kyunki effective particles ki number ghat jaati hai.
Osmotic pressure ka real-world use kya hai?
Osmotic pressure reverse osmosis (RO) filters me use hoti hai — jab solution par uske osmotic pressure se zyada pressure lagayi jaati hai, to solvent (pure water) ulti direction me membrane cross karta hai, jisse desalination aur water purification hoti hai.
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