Solutions Chemistry Notes PDF
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These notes cover various aspects of solutions in chemistry, including definitions, examples, and calculations. They discuss different types of solutions and factors affecting solubility. The document explores topics like boiling point elevation, freezing point depression, osmotic pressure, and ideal vs. non-ideal solutions.
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## CHP 2 SOLUTIONS **What is solution?** Explain the term solute and solvent. A homogeneous mixture of two or more than one component. * Regular arrangement (sb mix ho jata hai) **A solution having three components are called as** * QUATERNARY * BINARY * SINGLE * TERNARY | Solute | Solvent | Exa...
## CHP 2 SOLUTIONS **What is solution?** Explain the term solute and solvent. A homogeneous mixture of two or more than one component. * Regular arrangement (sb mix ho jata hai) **A solution having three components are called as** * QUATERNARY * BINARY * SINGLE * TERNARY | Solute | Solvent | Example | |---|---|---| | Solid | Liquid | Sea water, benzoic acid in water | | Solid | Solid | Metal alloys such as brass, bronze | | Solid | Gas | Iodine in air | # ABHISHEK SIR CHEMISTI? | LIQUID | LIQUID | Gasoline, ethanol in water | | LIQUID | SOLID | Amalgams of mercury | | LIQUID | GAS | chloroform | | GAS | GAS | air \[O<sub>2</sub>, N<sub>2</sub>, Ar] | | GAS | SOLID | H<sub>2</sub> in palladium | | GAS | LIQUID | carbonated water | **Which of the following is the correct example of solid solution in which solute is in gas phase?** * copper dissolved in gold * camphor in nitrogen gas * hydrogen in palladium * none of these # CAPACITY OF SOLUTION TO DISSOLVE SOLUTE :- | UNSATURATED | SATURATED | SUPERSATURATED | |---|---|---| | capacity se kam space occupy ho hai koh solute | capacity jya hai para occupy kr dena | capacity se jyade solute occupy krma | # ABHISHEK SIR CHEMISTI? **SOLUBILITY :-** The solubility of a solute is its amount per unit volume of saturated solution at a specific temperature. **UNIT** : mol L<sup>-1</sup> **Solubility** : **Molarity** : **no. of particle of solute / volume of soln in litre** # FACTORS AFFECTING SOLUBILITY :- * Nature of solute & solvent * Effect of temperature on solubility * Effect of pressure on solubility # NATURE OF SOLUTE & SOLVENT :- Like dissolves like. * Solute-Solute * Solvent-solvent * Solute - Solvent **How does the solubility of a gas in a water variis with temperature?** Gib doh chizzo ke beech attraction accha hai aur agar beech mai kuch temp laga de toh voh weak hoke bond break hoh jayega Temperature increases, Solubility decreases. # ASC # EFFECT OF PRESSURE ON SOLUBILITY :- **HENRYS LAW** It states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution. S = KHP S = KH (P) KH -> Henrys constant UNIT : mol L<sup>-1</sup> = KH mol<sup>-1</sup> Bar<sup>-1</sup> = KH # EXCEPTION'S OF HENRY LAW :- Gases like NH<sub>3</sub> and CO<sub>2</sub> do not obey Henrys law. The reason is that gases react with water. NH<sub>3</sub> + H<sub>2</sub>O = NH<sub>4</sub><sup>+</sup> + OH<sup>-</sup> CO<sub>2</sub> + H<sub>2</sub>O = H<sub>2</sub>CO<sub>3</sub> * inki solubility jyada hai as compared to Henrys laws # VOLATILE & NON-VOLATILE * aisa liquid joh 100℃ te temp se kam mai evapourate hon * aisa liquid job 100℃ ke temp se jyada mai evapourate hon # ASC # VAPOUR PRESSURE **The pressure created by vapours or the liquid is called as vapour pressure.** * jb paani vapour hoke upr jaata hai and vapour retürn aoke surface of liquid pe pressure deta hai usko hi vapour pressure. # RADULTS LAW Raoults law says that, “The Partial vapour Pressure of any volatile component of a solution is equal to the vapour pressure of the pure component (multiplied by its mole fraction in the solution.” P<sub>A</sub> = P°<sub>A</sub> . X<sub>A</sub> -> mole fraction Partial vapour pressure Let P<sub>A</sub> = partial vapour pressure of A. P°<sub>A</sub> = pure vapour pressure of A P<sub>B</sub> = partial vapour pressure of B P°<sub>B</sub> = pure vapour pressure of B PT = P<sub>A</sub> + P<sub>B</sub> = P°<sub>A</sub>. X<sub>A</sub> + P°<sub>B</sub>. X<sub>B</sub> X<sub>A</sub> + X<sub>B</sub> = 1 X<sub>B</sub> = 1-X<sub>B</sub> PT = P°<sub>A</sub>. X<sub>A</sub> + P°<sub>B</sub> (1-X<sub>A</sub>) PT = P°<sub>A</sub>. X<sub>A</sub> + P°<sub>B</sub> (1-X<sub>A</sub>) -> RAOUTS LAW... # GRAPH OF VARIATION OF VAPOUR PRESSURE WITH MOLE FRACTION OF SOLUTE :- * joh point upr se jah rhi hai voh positive * joh points niche se jah shi hai voh negative. **IDEA SOLD:** * Obeys Raoults law * No heat evolved or absorbed * Volume change nhi hoga eg: solate-solute / solvent - solvent comparable. # NON-IDEAL SOLUTION :- Jon ideal soin ki properties naah follow kne. **What is colligative propecities?** The physical properties of a solution that depends on the number of the solute particles in solution and not on thein nature is called as colligative properties. # RELATIVE LOWERING IN VAPOUR PRESSURE **When non-volatile solute is dissolve in volatile solvent its Vapour Pressure decreases** Hence, * P<sub>A</sub> = P°<sub>A</sub>. X<sub>A</sub> (volatile) * P<sub>B</sub> = P°<sub>B</sub>. X<sub>B</sub> (non-volatile) * PT = P<sub>A</sub> + P<sub>B</sub> * PT = P<sub>A</sub>. X<sub>A</sub> + P°<sub>B</sub>. X<sub>B</sub>) * PT = P°<sub>A</sub>. X<sub>A</sub> + * X<sub>A</sub> + X<sub>B </sub> = 1 * X<sub>A</sub> = 1-X<sub>B</sub> * ∴ PT = P<sub>A</sub> (1-X<sub>B</sub>) * ∴ P<sub>T</sub> = P°<sub>A</sub> - P°<sub>A</sub>. X<sub>B</sub> * P<sub>A</sub>. X<sub>B</sub> = P°<sub>A</sub> - PT * X<sub>B</sub> = (P°<sub>A</sub> - PT) / P°<sub>A</sub> Relative lowering in vapour pressure = (P°<sub>A</sub> - PT) / P°<sub>A</sub> * X = n<sub>2</sub> / (n<sub>1</sub> + n<sub>2</sub>) MOLE FRACTION 1 -> SOLVENT 2 -> SOLUTE # Relative lowering of vapour pressure :- X<sub>B</sub> = (P°<sub>A</sub> - PT) / P°<sub>A</sub> X<sub>B</sub> = (P°<sub>A</sub> - PT) / P°<sub>A</sub> X<sub>B</sub> = ∆P/ P°<sub>A</sub> # MOLAR MASS OF SOLUTE FROM VAPOUR PRESSURE LOWERIA * X<sub>B</sub> = P°<sub>A</sub> - PT / P°<sub>A</sub> * X<sub>B </sub>= ∆P / P°<sub>A</sub> * X<sub>B</sub> = n<sub>2</sub> / (n<sub>1</sub> + n<sub>2</sub>) where, n<sub>1</sub> + n<sub>2</sub> >> n<sub>1</sub> hi + h<sub>2</sub> = 1 * X = n<sub>2</sub> / n<sub>1</sub> * n<sub>2</sub> = W<sub>2</sub> / M<sub>2</sub> , n<sub>1</sub> = W<sub>1</sub> /M<sub>1</sub> * X = n<sub>2</sub> / n<sub>1</sub> * n<sub>2</sub> = W<sub>2</sub> / M<sub>2</sub> * n<sub>2</sub> = ∆P / P°<sub>A</sub> * W<sub>2</sub> / M<sub>2</sub> = ∆P / P°<sub>A</sub> * W2. Mi / W1.M2 = ∆P / P°<sub>A</sub> * M<sub>2</sub> = 1000 KbW<sub>2</sub> / ∆TbW<sub>1</sub> # DEPRESSION IN FREEZING POINT When non-volatite solute is dissolved in volatile soluent its vapour pressive decreases thus, fuweging point decreases. * The difference between the fuweging point of pure Solvent and that of the solution is called depression in freezing point. * ∆Tf = T°<sub>f</sub> - Tf **Define cryoscopic contant ? Write its unit ?** The fuweging point depression (∆Tf) is directly proportional to the molality of the solution. * ∆Tf α K<sub>f</sub> . m -> molality * ∆Tf = K<sub>f</sub> . m * Freezing Point * Cryoscopic constant The proportionality constant K<sub>f</sub> is called as fuweging point depression or cryoscopic constant. * ∆Tf = K<sub>f</sub> . m * K<sub>f</sub> = ∆Tf / m * mol/kg = K<sub>f</sub> * K mol kg<sup>-1</sup> = K<sub>f</sub> * Unit of cryoscopic constant. **Derive the relation between molar mass of freezing point and molar mass of solute?** The molality of solution is given by m = 1000W<sub>2</sub> / M<sub>2</sub> Wi * ∆Tf = K<sub>f</sub> 1000W<sub>2</sub> / M<sub>2</sub> Wi * M<sub>2</sub> = 1000K<sub>f</sub>W<sub>2</sub> / ∆TfW<sub>1</sub> # OSMOSIS The flow of Solvent mole cules fuom lower concent ration to higher concentration. # OSMOTIC PRESSURE The minimum perussure which must be applied to the solution side to prevent the osmosis. # TYPES OF SOLUTION 1) ISOTONIC SOLUTION * Two or more solutions having the same osmotic presswee are said to isotonic solutions. 2) HYPERTONIC SOLUTION : * If two solutions have unequal osmotic presswee the more concentrationed solution with higher osmotic perussure is said to be hypertonic solutions. 3) HYPOTONIC SOLUTION:- * The move dilutated solution exhibiting lower osmotion preussure is called as hypotonic solvcan. # OSMOTIC PRESSURE & CONCENTRATION OF SOLO Molarity = Do. of mass of solute / volume of soin in litre M = D<sub>2</sub> / V * molarity = no. of moles of solute / volume of soln in litre * π = MRT * π = n<sub>2</sub>RT / V # MOLAR MASS OF SOLUTE FROM OSMOTIC PRESSURE We know that π = n<sub>2</sub>RT / V But n<sub>2</sub> = W<sub>2</sub> / M<sub>2</sub> π = W<sub>2</sub>RT / M<sub>2</sub>V M<sub>2</sub> = W<sub>2</sub>RT/ πV # REVERSE OSMOSIS :- The pure solvent that fesus fuom solution into the pure solvent through semipermeable membrane is called as ewerse osmosis. # SEMIPERMEABLE MEMBRANE:- * It is a film such as cellophane which has pores large enough to allow the solvent mole alles to pass through them. These pores are small enough for not to allow the passage of larger solute molecules or ions of high mole cule mass through them. * no ke alpend korta # COLLIGATIVE PROPERTIES OF ELECTROLYTE 1) VANT'S HOFF FACTOR (i) It is defined as the ratio of the colligative property of a solution of electrolyte to the colligative property of a non-electrolyte solution of the same concentra- tion. * It is denoted by (i) * i = colligative property of electrolyte soln / colligative property of non-electrolyte soln * (OTF) = (ΔΤf)<sub>0</sub> (i) (OTb) = (ΔΤb)<sub>0</sub> (i) (ΔP)<sub>0</sub> (i) 2) MODIFICATION OF VANT HOFF FACTOR :- * OP = iπ<sub>0</sub> = iW<sub>2</sub>M<sub>1</sub>π<sub>1</sub>P<sub>2</sub> / M<sub>2</sub>W<sub>1</sub> * ∆Tb = iK<sub>b</sub>m = i 1000 KbW<sub>2</sub> / M<sub>2</sub>W<sub>1</sub> * ∆Tf = iK<sub>f</sub>m = i 1000 K<sub>f</sub>W<sub>2</sub> / M<sub>2</sub>W<sub>1</sub> * π = iMRT = iW<sub>2</sub>RT / M<sub>2</sub>V # BELATION BET VANTS HOEF FACTOR and DEGREE OF DISSOCIATION :- Considue an electrolyte Ax By that dissociates in aqueous solution as AxBy -> xA<sup>+</sup> + yB<sup>-</sup> Initially : 1 mol At equilibrium, 1 - d xd yd Total moles of dissociation = (1-d) + xd + yd = 1 + d (x+y-1) we know that (x+y) = D = 1 + d (n -1) i = 1 + d (n -1) i -1 = d (n -1) d = (i-1) / (n-1) ABHISHEK SIR CHEMISTRY YOUTUBE WISHING YOU ALL THE VERY BEST LIKE, SHARE & SUBSCRIBE
