Class 12 Chemistry Solutions PDF

SOLUTIONS Class by YAKSHU Introduction Matter is made up different kinds of particles and chemically the nature of these particles could be (a) Pure Substances, or (b) Mixtures These could be homogeneous or heterogeneous, the homogeneous mixture of more than one pure substance is known as SOLUTION....

SOLUTIONS Class by YAKSHU Introduction Matter is made up different kinds of particles and chemically the nature of these particles could be (a) Pure Substances, or (b) Mixtures These could be homogeneous or heterogeneous, the homogeneous mixture of more than one pure substance is known as SOLUTION. Every solution is made up of two components Solute and Solvent. If the physical state of solute and solvent are the same the component of lower mass is solute and component with higher mass is solvent. The solvent is one but the solute can be many, and such solutions containing only one solute are known as Binary Solution. 1 What is Solution 2 Expressing Concentration in Different Terms 3 Solubility of Gases and Solids in Liquids 4 Vapour Pressure of Liquid-Liquid & Raoult's Law for Volatile Solute 5 Vapour Pressure of Solid-Liquid Solutions & Raoult's Law for Non- Volatile Solute 6 Ideal & Non-Ideal Solutions 7 Colligative Properties and Determination of Molecular Mass 8 Abnormal Molar Masses 9 Van't Hoff Factor What is SOLUTION ?? Ek solution ko agar hum define kare toh solution hamara wo homogenous (ek tarah ke) mixture hota h jo 2 or 2 se jyada substances se bante h jinki compositions aur properties almost uniform rahti h. Jo substance solution banate h unko hum components bolte h. Ab solution meh kitne components present h uske basis par hum solutions ko binary (two components), ternary (three components), quaternary (four components) solution bolte h. Ab hum agar binary solution consider kare toh isme ek component hamara solvent aur ek component solute hota h. Jisme agar solute aur solvent components ki physical state same h tab jis component ka mass kam h wo solute aur jis component ka mass jyada h wo solvent hota h. Ab inn solutions ki daily life meh importance kya h usko hum solution ki different concentration ke basis par dekhte h. For example “(i) Jaise ki agar fluoride ion ki water meh 1 part per millions (ppm) concentration h toh ye tooth decay na ho usme help karta h. “(ii) Similarly, Brass - copper aur zinc ka homogeneous mixture h, German silver - copper, zinc aur nickel ka mixture h aur similarly bronze - copper aur tin ka mixture ab teeno mixture meh copper common h, lekin teeno mixtures ki properties different h isliye inke uses b alag hote h. Figure 1 : Depending upon physical state of components, solution may be of following types Iss table meh solid in liquid, liquid in liquid, aur gas in liquid bahut common solutions h aur generally hamara solvent kis physical state meh h uske basis par hum solutions ko classify karte h. Similarly jis solution meh water solvent hota h usko hum aqueous solution bolte h aur jis solution meh water solvent nahi hota usko hum non- aqueous solution bolte h aur non-aqueous solution meh solvent ether, benzene, carbon tetrachloride etc., hote h. POINT 1 Solution : A solution is a homogenous mixture of two or more chemically non- reacting substances whose composition can be varied within certain limits. Solvent : A solvent is that component of the solution which is present in larger amounts by mass. Solute : A solute is that component of the solution which is present in lesser amounts by mass. Expressing Concentration of Solutions : Concentration ko agar hum generally define kare toh concentration wo amount of solute h jo ek known amount of solvent meh dissolve kia jata h. Ab concentration ko hum different ways meh define kar sakte h : (i) Mass percentage or weight percentage (w / w % ) (ii) Volume percentage (v / v % ) (iii) Mass by Volume percentage (w / v % ) (iv) Molarity (M) (v) Molality (m) (vi) Mole fraction (χ) (vii) Normality (N) (i) Mass percentage or weight percentage (w / w % ) : Mass % of component ko calculate karne ke liye hum mass of a component per 100 g of the solution meh consider karte h. Jiske according mass of the component in the solution Mass % of component = × 100 total mass of the solution Ab isme agar hum mass of component A ko w A aur mass of component B ko w B consider kare tab humara formula hoga.  wA (w / w) % = × 100..........(1) wA + wB Isko hum (w / w % ) se represent karte h. SOLVED EXAMPLES Question : Define mass percentage or weight percentage and Calculate mass percent of solute of a solution which is prepared by adding 2 g of substance A to 18 g of water. Answer : It is defined as the weight of solute in grams per 100 g of solution. It is represented as (w / w % ). mass of the solute in the solution Mass % of solute = × 100 total mass of the solution ( ) Consider numerical Given : mass of solute w A = 2 g ( ) Mass of water w B = 18 g Mass of solution = (2 + 18) g = 20 g Now put all values in (w / w % ) formula, it becomes [ ] mass of the solute in the solution w A Mass % of solute = × 100 total mass of the solution [w A + w B ] 2g Mass % of solute = × 100 20g Mass % of solute = 10 % (ii) Volume percentage (v / v % ) : Volume % of component ko calculate karne ke liye hum volume of a component per 100 parts by volume of the solution consider karte h. Jiske according Volume % of component = volume of the component in the solution × 100 total volume of the solution Ab isme agar hum volume of component A ko v A aur volume of component B ko v B consider kare tab humara formula hoga. vA (v / v) % = × 100..........(2) vA + vB Isko hum (v / v % ) se represent karte h. SOLVED EXAMPLES Question : Define volume percentage and Calculate volume percent of solute of a solution which is prepared by adding 10 mL of ethanol to 100 mL of solution. Answer : It is defined as the volume of solute per 100 mL of solution. It is represented as (v / v) %. Volume of the solute in the solution Volume % of solute = × 100 total volume of the solution ( ) Consider numerical Given : volume of solute v A = 10 mL ( ) Volume of water v B = 90 mL Volume of solution = 100 mL Now put all values in (v / v) % formula, it becomes  [ ] volume of the solute in the solution v A Volume % of solute = × 100 total volume of the solution [v A + v B ] 10mL Volume % of solute = × 100 100mL Volume % of solute = 10 % (iii) Mass by Volume percentage (w / v % ) : Mass by volume % of a component ko calculate karne ke liye hum mass of solute ko 100 mL by volume of the solution meh consider karte h. Jiske according mass of the component in the solution Mass/Volume % of component = × 100 total volume of the solution in mL Ab isme agar hum mass of component ko w aur volume of component v consider kare tab humara formula hoga. w × 100 (w / v) % =..........(3) v(mL) Isko hum (w / v % ) se represent karte h. SOLVED EXAMPLES Question : Define mass by volume percentage and Calculate mass by volume percent of solute of a solution is prepared by adding 10 g of glucose to 100 mL of solution. Asnwer : It is defined as the weight of solute per 100 mL of solution. It is represented as (w / v) %.  mass of the solute in the solution Mass by Volume % of solute = × 100 total volume of the solution Given : mass of solute (w) = 10 g Volume of solution (v) = 100 mL Now put all values in (w / v) % formula, it becomes Mass by Volume % of solute = mass of the solute in the solution(w) × 100 total volume of the solution(v) 10g Mass by Volume % of solute = ✕100 100mL Mass by Volume % of solute = 10 % (iv) Molarity : Molarity of solution ko calculate karne ke liye kitne number of moles of solute per litre of solution meh dissolved hote h uska hum ratio consider karte h. Jiske according., number of moles of solute(mol) Molarity (M) = volume of solution in litres(L) Isko hum M se represent karte h aur iski units moles per litre (mol. L ) (equal to M) hoti h. Ab iss expression meh number of -1 moles of solute ko calculate karne ke liye hum mass of solute aur molar mass of solute ko consider karte h jiske according mass of solute(g) Number of moles of solute (mol) = molar mass of solute(g / mol) Ab inn dono expression ko sath meh consider kare tab humara molarity ka overall expression hoga.  mass of solute(g) Molarity (M) = molar mass of solute(g / mol) × volume of solution(L)..........(4) Ab agar volume of solution (mL) meh ho tab hum volume ko 1000 se divide karte h taki volume (L) meh convert ho jaye aur tab molarity ka overall expression hoga. Molarity (M) = mass of solute(g) × 1000(mL)..........(5) molar mass of solute(g / mol) × volume of solution(mL) × 1(L) SOLVED EXAMPLES Question : What is molarity ? Calculate molarity when 5 g NaOH dissolved in 2 L of solution. Answer : Molarity is defined as the no. of moles of solute present in 1 litre of solution. It is represented by M. It is expressed in mol / L units. mass of solute(g) Molarity = molar mass of solute(g / mol) × volume of solution in litres(L) Now consider numerical Given : mass of NaOH (solute) = 5 g Molar mass of NaOH (solute) = 40 g/mol Volume of solution = 2 L Now put all values in the formula, then it becomes 5g Molarity = 40g / mol × 2L 1 Molarity = or 0.0625 mol/L 16 Hello Points! Molarity ko hum (w / w) % aur (w / v) % se relate kar calculate kar sakte h jiske according humare pass 2 relation hote h. 10 × density × (w / w) % (1) Molarity = molar mass of solute 10 × (w / v) % (2) Molarity = molar mass of solute Similarly hum (w / w) % aur (w / v) % ko b ek expression ke through relate kar sakte h. (w / v) % = (density) × (w / w) % SOLVED EXAMPLES Question : Calculate molarity of NaOH solution whose (w / w) % = 80 and density = 1.2 g/mL. Answer : In this case, NaOH (w / w) % = 80 and density = 1.2 g/L are given and molar mass of NaOH = 40 g//mol. Thus, we are using here formula 10 × density × (w / w) % Molarity = molar mass of solute 10 × 1.2g / mL × 80 Molarity = 40g / mol Molarity = 24 M Along with this we can solve this numerical by using (w / w) % For this first we have to calculate number of moles of NaOH since (w / w) % = 80 It means 80 g NaOH present in 100 g solution and molar mass of NaOH = 40 g/mol. Thus, mass of NaOH(m) Number of moles (n) = molar mass of NaOH(M) 80g (n) = = 2 mol 40g / mol Also, density = 1.2 g/mL and mass of solution = 100 g, with the help of this we can calculate the volume of solution. mass density = volume 100g 1.2g / mL = volume 100g Volume = = 83.3mL 1.2g / mL Now we can calculate molarity by using its direct formula number of moles of solute(mol) × 1000(mL) Molarity = volume of solution(mL) × 1(L) 2mol × 1000mL Molarity = 83.3mL × 1L Molarity = 24 M (v) Molality : Molality of solution ko calculate karne ke liye kitne number of moles of solute per kilogram (or 1000 g) solution meh dissolved hote h uska hum ratio consider karte h. Jiske according., number of moles of solute(mol) Molality (m) = mass of solvent(kg) Isko hum m se represent karte h aur iski units moles per kg (mol. kg ) (equal to m) hoti h. Ab iss expression meh number of -1 moles of solute ko calculate karne ke liye hum mass of solute aur molar mass of solute ko consider karte h jiske according mass of solute(g) Number of moles of solute (mol) = molar mass of solute(g / mol) Ab inn dono expression ko sath meh consider kare tab humara molality ka overall expression hoga. mass of solute(g) Molality (m) =.......... molar mass of solute(g / mol) × mass of solvent(kg) (6) Ab agar mass of solvent (g) meh ho tab hum isko 1000 se divide karte h taki mass of solvent (kg) meh convert ho jaye aur tab molality ka overall expression hoga. mass of solute(g) × 1000(g) Molality (m) = molar mass of solute(g / mol) × mass of solvent(g) × 1(kg)..........(7) SOLVED EXAMPLES Question : Define molality. Calculate molality of 20 % (w / w) NaOH solution. Answer : Molality is defined as the no. of moles of solute present in 1 kg of solvent. It is represented by m and it is expressed as mol / kg. mass of solute(g) Molality (m) = molar mass of solute(g / mol) × mass of solvent(kg) Now consider the numerical Given : (w / w) % of NaOH = 20 This means 20 g of NaOH present in 100 g of solution with the help of this we can calculate number of moles of NaOH, since Molar mass of NaOH = 40 g/mol. Thus, mass of NaOH number of moles (n) = molar mass of NaOH 20g 1 n= = mol or 0.5 mol 40g / mol 2 Although mass of solution = 100 g which means mass of solvent = mass of solution - mass of NaOH mass of solvent = 100 - 20 = 80g Now we can calculate molality by using number of moles of solute(mol) × 1000(g) Molality (m) = mass of solvent(g) × 1(kg) 0.5mol × 1000g Molality (m) = 80g × 1kg 50 Molality (m) = or 6.25 m 8 Hello Points! Ek expression ke through hum molarity aur molality ko relate kar sakte. Jiske according 1000 × Molarity Molality (m) = 1000 × density - Molarity × Molar mass of solute SOLVED EXAMPLES Question : Calculate molality of 2 M NaOH solution (density = 1.2 g/mL). Answer : In this case molarity = 2 M and density = 1.2 g/mL are given and molar mass of NaOH = 40 g/mol. Thus, for calculating molality we can consider formula 1000 × Molarity Molality (m) = 1000 × density - Molarity × Molar mass of solute  1000 × 2M Molality (m) = 1000 × 1.2g / mL - 2 × 40g / mol m = 1.78 m Along with this we can solve this numerical by using molarity of NaOH Since molarity of NaOH = 2 M which means 2 moles of NaOH present in 1 L or 1000 mL solution. 1 mole of NaOH = 40 g which means 2 moles = 80 g (mass of solute) Also we have density = 1.2 g/mL so by using mass Density = we can calculate mass of the solution. volume Thus, mass 1.2(g / mL) = 1000(mL) Mass = 1.2(g / mL) × 1000 (mL) Mass of solution = 1200 g Mass of solvent = mass of solution - mass of solute Mass of solvent = 1200 - 80 = 1120 g Now we can calculate molality by using number of moles(mol) × 1000(g) Molality (m) = mass of solvent(g) × 1(kg) 2(mol) × 1000(g) Molality (m) = 1120(g) × 1(kg) Molality (m) = 1.78 m (vi) Mole fraction : Ek solution meh mole fraction ko calculate karne ke liye hum number of moles of one component aur total number of moles of all components (solute aur solvent) ka ratio consider karte h. Jiske according Mole fraction of a component = number of moles of a component Total number of moles of all components Isko hum chi (χ) se represent karte h aur iski koi b units nahi hoti. Isme agar hum ek binary solution ke case meh mole fraction of solute aur mole fraction of solvent consider kare toh uske according., Mole fraction of solute ( ) number of moles of a solute n A ( ) χA =..........(8) Total number of moles of solute & solvent (n A + n B ) Mole fraction of solvent ( ) number of moles of a solvent n B ( ) χB =..........(9) Total number of moles of solute & solvent (n A + n B ) Sabhi component ke mole fraction ko agar hum add kar de toh wo equal to one hota h. nA nB χA + χB = + = 1..........(10) nA + nB nA + nB Similarly agar hume ek binary solution meh ek component ka mole fraction dia ho tab hum dusre component ka mole fraction calculate kar sakte h. χ A = 1 - χ B or χ B = 1 - χ A SOLVED EXAMPLES Question : Define mole fraction. Calculate mole fraction of 2 mol NaOH in 10 mol water. Answer : Mole fraction is the ratio of number of moles of one component to the total number of moles (solute and solvent) present in the solution. It is represented as chi (χ) and it is a dimensionless quantity. For example if consider a binary solution in which if number of moles of solute (A) = n A and number of moles of solvent (B) = n B Then, Mole fraction of component A ( ) number of moles of component A n A ( ) χA = Total number of moles of all components (n A + n B ) Mole fraction of component B ( ) number of moles of component B n B ( ) χB = Total number of moles of all components (n A + n B ) Also, sum of mole fraction of both the components nA nB χA + χB = + =1 nA + nB nA + nB Now consider numerical according to which ( ) Mole fraction of NaOH n A = 2 mol Mole fraction of water (n B ) = 10 mol Total moles of solution (n A + n B ) = 12 mol Thus, mole fraction of both components can be calculated as, Mole fraction of component A ( ) number of moles of component A n A (χ A ) = Total number of moles of all components (n A + n B ) 2 1 ( A ) 12 6 χ = = Mole fraction of component B ( ) number of moles of component B n B (χ B ) = Total number of moles of all components (n A + n B ) 10 5 ( A ) 12 = 6 χ = (vii) Normality : Normality of solution ko calculate karne ke liye kitne number of gram equivalents of solute per litre of solution meh dissolved hote h uska hum ratio consider karte h. Jiske according number of gram equivalents of solute(g. eq. ) Normality (N) = volume of solution(L) Isko hum N se represent karte h aur iski units gram equivalent per ( ) litre g. equiL - 1 (equal to N) hoti h. Ab iss expression meh number of gram equivalents of solute ko calculate karne ke liye hum mass of solute aur equivalent mass of solute ko consider karte h jiske according mass of solute Number of gram equivalents of solute = equivalent mass of solute..........(11)  molar mass Equivalent mass of solute =..........(12) ( ) n-factor n f Ab inn dono expression ko sath meh consider kare tab humara normality ka overall expression hoga. mass of solute Normality (N) =.......... equivalent mass of solute × volume of solution(L) (13) Ab agar volume of solution (mL) meh ho tab hum volume ko 1000 se divide karte h taki volume (L) meh convert ho jaye aur tab normality ka overall expression hoga. Normality (N) = mass of solute(g) × 1000(mL).......... equivalent mass of solute (g/equiv.) × volume of solution(mL) × 1(L) (14) Ab equivalent weight nikalne ke liye hume n factor n f pata hona ( ) chaiye so toh sab se pehle n factor kya hota h. So n factor (nf ) basically acidic compounds ki acidity matlab ye [ ] kitne H + ions release kar sakta h aur basic compounds ki basicity [ ] ions release kar sakta h aur ye salt ke case matlab ye kitne OH - meh total charge jo cation par present ho wo hota h. For example in case of acidic compounds like ( ) [ ] (i) HCl iska n factor n f = 1 hoga kyunki ye 1 H + ion release karta h, ( ) [ ] ions release (ii) H 2SO 4 iska n factor n f = 2 hoga kyunki ye 2 H + karta h, ( ) [ ] (iii) H 3PO 4 iska n factor n f = 3 hoga kyunki ye 3 H + ions release karta h. For example in case of basic compounds like ( ) (i) NaOH iska n factor n f = 1 hoga kyunki ye 1 OH -[ ] ion release karta h, (ii) Ca(OH) 2 iska n factor (nf ) = 2 hoga kyunki ye 2 [OH - ] ions release karta h, (iii) Al(OH) 3 iska n factor (nf ) = 3 hoga kyunki ye 3 [OH - ] ions release karta h. Similarly in case of salts ( ) (i) NaCl iska n factor n f = 1 hoga kyunki ye 1 Na +[ ] ion release karta h aur iss ke case meh charge 1 h. ( ) [ ] (ii) CaCl 2 iska n factor n f = 2 hoga kyunki ye 1 Ca + 2 ion release karta h aur iss ke case meh overall charge 2 h. SOLVED EXAMPLES Question : Define normality. Calculate normality of solution of sulfuric acid which is prepared by dissolving 0.49 g of acid in 250 cm 3 of solution. Answer : The normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre of the solution. It is represented by the symbol, N and its unit is g. equiL - 1 mass of solute Normality (N) = equivalent mass of solute × volume of solution(L) Now consider numerical Given : mass of H 2SO 4 = 0.49 g Volume of solution = 250 cm 3 or 250 mL because 1cm 3 = 1mL( ) For equivalent mass of H 2SO 4 Its molar mass = 98 g/mol and its n factor n f = 2( ) molar mass Thus, its equivalent mass = ( ) n-factor n f 98(g / mol) equivalent mass = = 49g / equiv 2 Now we can calculate its normality by using Normality (N) = mass of solute(g) × 1000(mL) equivalent mass of solute(g / equiv) × volume of solution(mL) × 1(L) 0.49g × 1000mL Normality (N) = 49g / equiv × 250mL × 1L N = 0.04N Hello Points! n-factor n f is the basicity of an acid or acidity of a base 1. Acidity of a base - It is given by the Number of replacable of OH - ions For example - NaOH has 1 replacable OH - ion hence n f = 1 Ca(OH) 2 has 2 replacable OH - ions hence n f = 2 Al(OH) 3 has three replacable OH - ions hence n f = 3 2. Basicity of an acid : It is the number of replaceable H + ions For example, HCl is monobasic acid, n f = 1, as one H + ion is replaceable. H 2SO 4, is dibasic acid i.e. n f = 2 as two H + ions are replaceable. H 3PO 4, (phosphoric acid) is tribasic acid i.e. n f = 3 as three H + ions are replaceable. H 3PO 3 (phosphorus or phosphonic) is dibasic, i,e, n f = 2 as there are 2 H + replacable ions 3. n factor of salt = Total charge present on cation For example - NaCl has n f = 1 CaCl 2 has n f = 2 Ek expression ke through hum normality aur molarity ko relate kar sakte h. Jiske according Normality (N) = molarity(M) × n-factor n f ( ) SOLVED EXAMPLES Question : Calculate the number of oxalic acid molecules in 100 mL of 0.02 N oxalic acid solution. Given : normality of solution = 0.02 N Answer : Volume of solution = 100 mL According to relationship between molarity and normality, we can say Normality (N) = molarity(M) × n-factor n f ( ) ( ) For oxalic acid n factor n f = 2 since it gives two H + [ ] ions, now putting the values in the above formula. It becomes ( ) Normality (N) = molarity(M) × n-factor n f 0.02N = molarity(M) × 2 Molarity (M) = 0.01M Now according to molarity definition number of moles of solute(mol) × 1000(mL) Molarity = volume of solution(mL) × 1(L) 0.01 × 100mL number of moles of solute (mol) = 1000mL - number of moles of solute = 1 × 10 mol 3 Now 1mole = 6.023 × 10 23 molecules 1 × 10 - 3mol = 1 × 10 - 3 × 6.023 × 10 23molecules = 6.023 × 10 20molecules Therefore, the number of oxalic acid molecules in 100 mL of 0.02 N oxalic acid solution is 6.023 × 10 20 molecules. Solubility of Gases and Solids in Liquids : Solubility ko agar hum define kare toh iske according kisi substance ka wo maximum amount jo ek specific temperature par ek specific amount of solvent meh dissolve ho usko hum solubility kahte h aur kisi b substance ki solubility nature of solute, nature of solvent, temperature aur pressure par depend karti hai. Ab inn effects ko hum 2 ways meh study kar sakte h. (1.) Solubility of solid in liquids (2.) Solubility of gas in liquids POINT 2 Solubility of a substance expresses the maximum amount of it which can be dissolved in a specific amount of solvent at a specific temperature. It depends upon the nature of solute, solvent, temperature and pressure. (1.) Solubility of solid in liquids : (i) Dissolution : Jab kabhi b hum kisi solid solute ko ek liquid solvent meh continuously add karte h tab uss solid ki concentration toh increase hoti h aur wo solute continuously uss solvent meh dissolve b ho rha h tab uss process ko hum dissolution bolte h. (ii) Crystallization : Lekin ek stage ke baad solid solute uss solvent meh uss particular temperature par dissolved nahi hota kyunki tab uss solution meh jo solute particles already present h aur solute particles hum bahar se add kar rahe h wo aapas meh collide karte h aur precipitate ki form meh bahar nikalte h tab uss process ko hum crytsallization bolte h. Iss time par rate of dissolution aur rate of crystallization equal ho jata h matlab yahan ek equilibrium generate hoti h. Solute + solvent ⇌ solution Iss solution ko hum saturated solution b bolte h. Similarly, unsaturated solution wo solution hota h jisme maximum solute same temperature par dissolve hota h. Accordingly Such a solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution. An unsaturated solution is one in which more solute can be dissolved at the same temperature. The solution which is in dynamic equilibrium with undissolved solute is the saturated solution and contains the maximum amount of solute dissolved in a given amount of solvent. Thus, the concentration of solute in such a solution is its solubility. Factors affecting Solubility in Solids in Liquids : Solubility of solid in liquids generally 2 factors par depend karta h. (a) Nature of the solute and solvent (b) Temperature Solid in liquid ki solubility pressure par depends nahi karti. (a) Nature of the solute and solvent : Koi b solid kisi liquid meh dissolve tabhi hota h jab inn dono meh similar properties hoti h. Jaise ki humne pada b h like dissolves like. Toh iss statement ke basis par hum kah sakte h ki polar compound jaise NaCl, polar solvent jaise water meh completely dissolve ho sakta h aur non polar solvents jaise ether, benzene isme NaCl almost insoluble rahta h. Similarly, non polar compound jaise naphthalene, non polar solvent ether, benzene meh dissolve ho jata h aur polar solvent water meh dissolve nahi hota. Ab ionic compounds like NaCl, covalent compounds like sugar ke comparison water meh jyada dissolve hote h. Ab aisa kyun hota h? [ ] [ ] Sodium chloride meh Na + aur Cl - ion present hote h aur water ek polar compound h iska matlab inke beech meh strong [ ] electrostatic forces present h jisme negative ions Cl - ko solvent ka positive pole attract karta h aur similarly positive ions [Na ] ko + solvent ka negative pole attract karta h. Jis wajah se jo electrostatic forces present thi inke beech wo ab weak ho gyi h. Iska matlab ab [Na ] aur [Cl ] ion dubara combine kar NaCl nahi bana sakte + - matlab jo ions solution meh pehle free form meh present the wo ab hydrated form meh present h. Ab NaCl ko uske ions me todne k liye jo energy chahiye vo lattice energy hoti hai aur jab ye ions me tut jata hai to water molecule Na + and Cl - ko solvate kar dete hai jisko hydration bolte hai aur iss solvation process me bahut sari energy release hoti hai jo require lattice energy ko compensate karti hai. Ab koi substance water me tab soluble hota hai jab uski hydration energy lattice energy se jyada ( hoti hai. ΔH hyd > ΔH lattice ) (b) Effect of temperature : Solubility of solid in liquids in case of ionic compounds temperature par 2 tarike se depend karti h. (i) Solubility increases with increase in temperature : Kuch ionic compounds jaise NaNO 3, KNO 3, NaCl, KCl etc., inki solubility temperature increase karne par increase hoti h kyunki inka jab hum dissolution karwate h tab inko heat chaiye hoti h matlab iss case meh ye endothermic process hota h ΔH sol > 0( ) Solute + solvent + heat ⇌ Solution..........ΔH sol = + ve Isko agar hum Le Chatelier’s principle ke basis par dekhe toh jab hum temperature increase karte h tab reaction ki equilibrium uss direction meh jati h jisme heat absorb hoti h matlab forward direction, isliye jab hum temperature increase karte h tab jyada solute solvent meh dissolve hote h. (ii) Solubility decreases with increase in temperature : Kuch ionic compounds aise b hote h jinki solubility temperature increase karne par decrease hoti h kyunki inka jab hum dissolution karwate h tab ye heat release karte h matlab iss case meh ye exothermic process hota ( h ΔH sol < 0 ) Solute + solvent ⇌ Solution + heat..........ΔH sol = - ve For example Li 2CO 3, Na 2CO 3. H 2O etc., (1.) Solubility of Solid in Liquid - Solubility of solids in liquids is independent on pressure. It generally follows the principal of Like Dissolves Like that is polar will dissolve polar and non-polar compounds will dissolve non-polar compounds because solids and liquids are highly incompressible and and remain unaffected by change in pressure For Example : When we try to wash the butter off our hands by using just water it wont wash off because water is polar and butter is non polar. Solubility also depends on solute-solvent interactions.If solute solvent interaction is more then solubility will also be more. The solubility of a solid in liquid is significantly affected by temperature changes. In general the solubility of a saturated solution depends upon the enthalpy change. ΔH dissolution ΔH dissolution = ΔH Lattice Energy + ΔH Hydration Energy. (a) If the enthalpy of dissolution is negative i.e. it is an Exothermic process then the Solubility will Decrease on Increasing the Temperature. (b) If the enthalpy of dissolution is positive i.e. it is an Endothermic process then the Solubility will Increase on increasing the temperature because if we apply Le Chatelier's Principle the dissolution equilibrium will shift in forward direction. (2.) Solubility of gas in liquids : Almost sabhi gases water meh ek limit tak dissolve hoti h. Natural water meh b dissolved oxygen present hoti h jis wajah se aquatic life lakes, rivers, sea meh exist kar pa rahi h. Ab solubility ko gas in liquid ki terms meh define kare toh iske according Toh gas ki wo volume jo ek fixed volume of liquid meh ek particular temperature par dissolve ho usko hum solubility bolte h. Solubility of gas in liquid ko hum molarity aur mole fraction ki form meh b express kar sakte h. Factors affecting Solubility in Gas in Liquids : Solubility of gas in liquids generally 3 factors par depend karta h. (a) Nature of the solute and solvent (b) Temperature (c) Pressure (a) Nature of the solute and solvent : Gases jaise hydrogen, oxygen, nitrogen etc., ye water ke meh sirf limited amount meh dissolved hoti h kyunki ye gases water ke sath reaction easily show nahi karti aur gases jaise CO 2, HCl, NH 3 etc., ye water meh highly soluble hoti h kyunki ye gases water ke sath easily reaction show kar sakti h. (b) Temperature : Gas in liquid ki solubility temperature increase karne par decrease hoti h kyunki inka jab hum dissolution karwate h tab ye heat release karte h matlab iss case meh yeh exothermic ( process hota h ΔH sol < 0 ) Solute + solvent ⇌ Solution + heat..........ΔH sol = - ve Isko agar hum Le Chatelier’s principle ke basis par dekhe toh jab hum temperature increase karte h tab reaction ki equilibrium backward direction meh hoti h, isliye jab hum temperature increase karte h tab solute ki solvent meh solubility kam hoti h. (c) Effect of Pressure (Henry’s Law) : Similarly gas in liquid ki solubility pressure increase karne par increase karti h. Iske liye hum ek system consider karte h jisme gas liquid ke sath equilibrium meh h aur iss system ke lower part meh solution h aur upper part meh gaseous h aur iska pressure p and temperature T h. Ab kyunki gas aur liquid equilibrium meh h iska matlab jitne gas particles enter kar rahe h utne hi bahar b nikal rahe h, ya fir hum ye kah sakte h. Rate of dissolution = rate of evaporation Lekin jab hum system se jyada pressure apply karte h tab gas particles sukh (compressed) jate h, matlab gas ki volume kam ho jati h. Jis wajah se gas particles per unit volume increase ho jate h aur ab jyada se jyada gaseous molecules liquid ke surface par takrate(strike) h, jis wajah se jyada ab jyada gaseous molecules dissolve hote aur solubility pressure increase karne par increase hoti h. Figure : Effect of pressure on the solubility of a gas. The concentration of dissolved gas is proportional to the pressure on the gas above the solution Ab pressure increase karne par solubility of gas in liquid increase kyun karti h? So isko samajhne ke liye William Henry ne Henry law dia tha jiske according Constant temperature par equilibrium solution meh gas ka wo amount jo per unit volume of solvent meh dissolved hota h, wo pressure ke directly proportional hota h Isko mathematically dekhe so iske liye hum mass of gas dissolved in a unit volume of the solvent ko m consider karte h aur gas ke pressure ko p consider karte h m∝p m = K. p..........(15) Iss expression meh K proportionality constant aur iski value nature of the gas, nature of the solvent, temperature aur pressure ki units par depend karti h. Iss time period meh Dalton ne b ye explain kia tha ki solubility of gas in liquid partial pressure of the gas par depend karti h aur issi basis par agar hum solubility ko mole fraction ki terms meh consider kare tab mole fraction of gas partial pressure of gas ke directly proportional hota h χ∝p χ = K′. p 1 p= χ K′ ( p = K Hχ where K H = 1 K′ )..........(16) Here , K H is Henry’s law constant aur iski units atm or bar hoti h. Ab iss expression ke basis par agar hum partial pressure aur mole fraction ke beech plot consider kare tab hume ek straight line graph milta h jiska slope = K H (in atm or bar) hota h. Figure : Experimental results for the solubility of HCl gas in cyclohexane at 293 K. The slope of the line is the Henry’s Law constant K H Similarly agar hum different gases ki different temperature par K H value consider kare. Figure : Value of Henry’s Law constant for some selected gases in water Inn value se hum ye kah sakte h : (i) Henry’s law constant, K H nature of the gas par depend karta h. (ii) Kisi particular pressure par jitni jyada Henry’s law constant, K H ki value hogi utni hi kam gas in liquid ki solubility hogi. ∵ ( χ= 1 KH p ) (iii) Henry’s law constant, K H ki value temperature increase karne par increase karti h jiske basis par hum ye kah sakte h gas ki solubility temperature increase karne par decrease karti h. Ab Henry’s law sirf specific condition par hi follow hota h : (LIMITATIONS) (i) Jab pressure low aur temperature high hoga matlab jab gas almost ek ideal gas ki tarah kaam karegi tab Henry’s law follow hoga. (ii) Gas highly soluble nahi honi chahiye, iske sath gas solvent ke sath react b nahi karni chahiye aur solvent meh dissociation aur association b show nahi karni chaiye. Applications of Henry’s Law : (i) In the production of carbonated beverages : Cold drinks, soda water, beer etc. meh CO 2 ki solubility ko increase karne ke liye, bottles ko high pressure par sealed kia jata h. Jab ye bottle ko open kia jata h tab partial pressure of CO 2 decrease hota h jis wajah se solubility decrease hoti h aur CO 2 bottle se bahar jati h. (ii) In the deep sea living : Deep sea divers (Scuba divers) underwater compressed air par depend karte, compressed air meh N 2 aur O 2 present hoti h aur ye gases normal pressure par blood meh limited amount meh hi dissolved hoti h. Lekin jab scuba divers deep sea meh jaate h tab wahan pressure high hota h jiski wajah se N 2 gas blood meh jyada dissolve hoti h aur jab scuba diver wapis surface par aate h tab pressure decrease karta h jis wajah se jo N 2 blood meh dissolve hui thi wo bubbles banati hai jis wajah se blood flow ruk jata h aur ye further never impulses ko affect kar bends condition generate karta h. Ab bends ko avoid karne ke liye scuba divers air cylinders use karte h jisme 11.7 % He, 56.2 % nitrogen aur 32.1 % oxygen gases hoti h. (iii) In the function of lungs : Jab air lungs meh enter karti h tab partial pressure of oxygen high hota h aur jab ye oxygen haemoglobin ke sath combine karti h tab ye oxyhaemoglobin banati h. Lekin tissues meh partial pressure of oxygen low hota h jis wajah se oxyhaemoglobin se jo oxygen release hoti hai wahi cells ki functioning meh use hoti h. (iv) For climber or people living at high altitudes : ab high altitudes par partial pressure of oxygen low hota h jiske according jo public wahan rahti unke tissues or blood meh oxygen ki concentration low hoti h, jis wajah se wo kafi weak hote h aur unki thinking power b strong nahi hoti h jisko hum anoxia condition bolte h. POINT 4 Solubility in Terms of Gas in Liquid Volume of the gas (in cm 3 converted to S.T.P) that can dissolve in a fixed volume of the liquid to form the saturated solution at a given temperature. (2.) Solubility of Gas in Liquid - Almost all the gases are soluble in water, the solubility of Oxygen in water is the one of the major requirements of survival.The solubility depends on a number of factors such as : (a) Nature of gas : Easily liquefiable gases are more soluble in water, for example NH 3 more soluble than O 2. Gases which can react with water are more soluble, for example CO 2 + H 2O → H 2CO 3 SO 2 + H 2O → H 2SO 3 O 2 + H 2O → Less Soluble (b) Temperature : Dissolution of gases in water is an exothermic process so with increasing temperature the solubility decreases, this is because on heating the solution of gas, some gas is expelled out, thus Gas + Solvent ⇔ Solution + Heat Hence according to Le Chatelier’s Principle the increase in temperature would favour the backward direction and solubility would decrease. (c) Effect of Pressure : On increasing pressure number of collision of gas molecules increases so solubility also increases.The effect of pressure on solubility can be explained by the Henry’s Law: This law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution If we use the mole fraction of a gas in the solution as a measure of its solubility, then it can be said that the mole fraction of gas in the solution is proportional to the partial pressure of the gas on the solution. That means if partial pressure of gases over the solution increases then the mole fraction of gas in water also increases. p α χ dissolved p = K H × χ dissolved (Units of K H : atm or bar or pa ) p χ dissolved = KH ( ) therefore if K H increases then χ dissolved decreases and hence solubility of gas in solution decreases K H is the Henry’s Constant and its value depends upon the following factors : 1. Temperature – K H increases if Temperature increases. 2. Nature of Gas - Different gases have different K H values By conducting experiments and observing the results it has been concluded that as K H increases the Solubility decreases. Henry’s law finds several applications in industry and explains some biological phenomena. Notable applications among these are: To increase the solubility of CO 2 in soft drinks and soda water, the bottle is sealed under high pressure. Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen). At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia. Limitation Henry's Law Henry's Law is applicable if : 1. Pressure is low as at high pressure the law loses its accuracy and shows deviations. 2. Temperature is not very low. 3. The solubility of gas is low 4. The gas does not react chemically with the solvent. Vapour Pressure of Liquid Solutions : Liquid solution meh solvent liquid hota h aur solute gas, liquid or solid meh se koi b ho sakta h. Agar hum binary solution (2 components) consider kare toh isme gas in liquid toh hum already discuss kar chuke hain ab hum yahan solid in liquid and liquid in liquid consider karenge jisme jo property discuss karte h wo vapour pressure hoti h. Liquids in liquids solution meh dono components volatile hote h aur solids in liquids meh solute non-volatile and solvent volatile hota h. Ab vapour pressure kya hota h?? Jab hum kisi liquid ko ek closed vessel meh heat karte h tab wo liquid evaporate hokar vapour meh convert hota h aur uss vessel meh jo space h vapour usko occupy karta h. Ab kyunki liquid-vapour meh convert ho raha h matlab liquid ka amount decrease ho raha aur jaise jaise evaporation increase hota h liquid ka amount kam aur vapour ka amount increase hota h, lekin jaise vapour ka amount increase hota h wo uss vessel meh random movement show karta h aur iss random movement ki wajah se kuch vapour particles liquid ke surface par strike kar condensed hote h. Condensation process evaporation process ka opposite hota h matlab ab iss vessel meh evaporation aur condensation dono ek sath ho rahe h. Ab kuch time ke baad ek aise stage aayega jab rate of evaporation aur rate of condensation equal hoga matlab inn dono ke beech equilibrium established ho jayegi aur equilibrium par jo pressure vapour ki taraf se lgta husko hum vapour pressure bolte h. Factors on which vapour pressure of liquid depends : The vapour pressure of a liquid depends upon : (i) Nature of the liquid : Har liquid ka ek characteristic vapour pressure hota h kyunki har liquid meh different magnitude of intermolecular forces hoti h. Jis liquid meh weakest intermolecular force hoti h wo easily vapour meh convert ho jata h jis wajah se uss liquid ka vapour pressure high hota h. For example diethyl ether aur alcohol meh intermolecular forces water ke comparison kam hoti h isliye inka vapour pressure water ke comparison jyada hota h. (ii) Temperature : Liquid ka vapour pressure temperature increase karne par increase karta hai kyunki temperature increase karne par liquid molecules ki kinetic energy increase karti h jis wajah se jyada liquid molecules -vapour molecules meh convert hote h jis wajah se uss liquid ka vapour pressure high hota h. (iii) Adding non-volatile solute : jab hum ek non volatile solute ko solution meh dalte h tab solvent ka vapour pressure decrease ho jata h kyunki tab number of solvent molecules per unit area decrease ho jate h. POINT 5 Vapour pressure The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure. Vapour pressure and factors affecting vapour pressure Vapour pressure measures the escaping tendency of liquid molecules into vapour phase The vapour pressure is directly proportional to Volatility and inversely to Boiling Point The vapour pressure depends on a number of factors such as : (a) Temperature - As the temperature increases the vapour pressure also increases because with increase in temperature the Kinetic Energy of molecules also increases, hence they leave the liquid surface and become Vapour. (b) Nature of Liquid : Lesser the inter-molecular forces between molecules easier it is for the molecule to leave the surface and become vapour. This is the reason why the vapour pressure of ether is greater than ethanol as ethanol has hydrogen bonding present but not in ether. (c) Addition of non-volatile Solute : It decreases the vapour pressure of solvent, because the number of molecules of solvent per unit area decreases. Vapour Pressures of Liquid-Liquid Solutions and Raoult’s Law (Raoult’s Law for Volatile Solutes) Jab humare pass ek liquid in liquid ka solution hota h toh usme dono component volatile hote h. Ab agar inn components ko hum heat kare toh ye evaporate ho jate h jiski wajah se ek time ke baad liquid aur vapour meh equilibrium established hoti h. Equilibrium ke baad dono components se jo vapour pressure lagta h jisko hum partial pressure bolte h jiski value solution meh jo component present h uske mole fraction aur pure component ke vapour pressure par depend karta h. Jisko hum Raoult’s Law b bolte h. Isko agar hum mathematically consider kare so iske liye hum 2 completely miscible volatile liquids 1 and 2 ka mixture consider karte h jisme χ 1 and χ 2 inke mole fraction aur p 1 and p 2 inke partial vapour pressure, similarly p 1∘ and p 2∘ inke pure state ke vapour pressure ko represent karta h. Jiske according humara Raoult’s Law ka expression hoga : p 1 = χ 1. p 1∘ and p 2 = χ 2. p 2∘..........(17) Ab agar hum system ke total pressure ko P consider kare same temperature par so according to Dalton Law of partial pressure : P = p1 + p2 P = χ 1. p 1∘ + χ 2. p 2∘ ( ) P = 1 - χ 2. p 1∘ + χ 2. p 2∘ ( ) P = p 2∘ - p 1∘ χ 2 + p 1∘..........(18) Equation (18) se hum ye conclude kar sakte h : (i) Solution ka total vapour pressure ko hum ek component ke mole [ ] [ fraction se relate kar sakte h χ 1 = 1 - χ 2 or χ 2 = 1 - χ 1 ] (ii) Solution ka total vapour pressure mole fraction of component 2 se linear relation show karta h kyunki p 1∘ aur p 2∘ constant hote h. (iii) Pure component 1 and 2 ke vapour pressure ke basis par solution ka total pressure increase or decrease kar sakta h agar mole fraction of component 1 increase hoga. Isko agar hum vapour pressure aur mole fraction ke plot ke basis par consider kare Figure : The plot of vapour pressure and mole fraction of an ideal solution at constant temperature. The dashed lines I and II represent the partial pressure of the components. (It can be seen from the plot that p 1 and p 2 are directly proportional to χ 1 and χ 2 respectively). The total vapour pressure is given line marked III in the figure. Iss plot ke basis par hum ye kah sakte h components ke vapour pressure aur mole fraction aapas mein Linear functions h. (i) Jab χ 1 = 1 ho tab solution meh pure liquid 1 present hoga (curve I) p 1 = p 1∘ × 1 = p 1∘ (ii) Aur jab χ 1 = 0 ho tab solution meh pure liquid 2 present hoga. (curve II) p 1 = p 1∘ × 0 = 0 (iii) Total pressure (curve III) represent kar raha h jo Dalton law of partial pressure ke according component 1 and 2 ke partial pressure ka sum hota h. P = p1 + p2 P = χ 1. p 1∘ + χ 2. p 2∘ Ab hamare solution meh jo vapour present h wo liquid ke sath equilibrium meh h so uski composition hum component ke partial pressure se calculate kar sakte h. Jiske liye vapour phase meh component 1 and 2 ke mole fraction ko hum y 1 and y 2 consider karenge. Ab agar hum Dalton law of partial pressure consider kare so iske according Partial pressure of a component = Mole fraction of the component × total pressure in vapour phase p 1 = y 1 × p..........(19) p 2 = y 2 × p..........(20) Iske basis par hum general form, p i = y i × p total use kar sakte h. p1 Similarly, Mole fraction of component 1 in vapour phase y 1 = p  p2 Mole fraction of component 2 in vapour phase y 2 = p Iske basis par isko hum general form Mole fraction of a component in vapour phase = partial vapour pressure of a component / total vapour pressure Vapour Pressure of Liquid - Liquid Solutions Suppose there is binary solution of two volatile liquids and denote the two components as 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually an equilibrium would be established between vapour phase and the liquid phase. Let the total vapour pressure at this stage be p total and p 1 and p 2 be the partial vapour pressures of the two components 1 and 2 respectively. These partial pressures are related to the mole fractions χ 1 and χ 2 of the two components 1 and 2 respectively. Raoult gave the quantitative relationship between the partial pressure and mole fraction According to him for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Thus, for component 1 p1 ∝ χ1 p 1 = p 1∘ χ 1 p 1∘ is the vapour pressure of pure component 1 at the same temperature. Similarly, for component 2 p 2 = p 2∘ χ 2 p 2∘ is the vapour pressure of the pure component 2. According to Dalton’s law of partial pressures, the total pressure ( p total ) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as: p total = p 1 + p 2 p total = p 1∘ χ 1 + p 2∘ χ 2 = (1 – χ 2 ) p 1∘ + χ 2 p 2∘ = p 1∘ + (p 2∘ - p 1∘ ) χ 2 We can conclude some crucial points from the last equation : (i) Total vapour pressure over the solution can be related to the mole fraction of any one component. (ii) Total vapour pressure over the solution varies linearly with the mole fraction of component 2. (iii) Depending on the vapour pressures of the pure components 1 and 2, total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component 1. (iv) In the p - χ diagram shown χ 1 and χ 2 and p 1 and p 2 represent the mole fraction and partial pressure respectively of constituent present in liquid phase. The composition of vapour phase in equilibrium with the solution is determined by the partial pressure of the components If Y 1 and Y 2 are the mole fraction of components 1 and 2 respectively in the vapour phase then : p 1 = Y 1P Total or Y 1 = (p 1) / (PTotal ) and p2 = Y2PTotal or Y2 = (p2) / (PTotal ) Vapour Pressure of Solutions of Solids in Liquids (Raout’s Law for non-volatile solutes) Isko padhne se pehle hume ye pata hona chahiye ki jab hum ek non- volatile solute ko solution meh add karte h tab uss solution ki evaporation tendency kam ho jati h kyunki uss solution meh kuch solute particles liquid surface par solvent ki position ko occupy kar lete h jis wajah se uss solution ka vapour pressure decrease ho jata h. Raout’s Law for non-volatile solutes : Ab according to Raoult’s Law volatile component ka partial vapour pressure, mole fraction ke directly proportional h. Lekin jab solute non volatile hoga, tab sirf solvent molecules hi vapour phase meh present honge. Jis basis par hum ye conclude kar sakte h ki solution ka vapour pressure solvent ke vapour pressure ke equal hoga. Jiska matlab Vapour pressure of the solution = vapour pressure of the solvent in the solution Ab agar hum solvent ke vapour pressure ko p 1 aur mole fraction of solvent ko χ 1 consider kare, tab Raoult’s Law ke according solvent ka vapour pressure hoga p1 ∝ χ1 p 1 = p 1∘ χ 1 p = p 1∘ χ 1..........(21) p (solution) = p ∘ (pure solvent) × mole fraction of solvent Since p ∝ χ 1 or p = p 1∘ χ 1 p = χ1 p 1∘ Ab agar dono side ko 1 se subtract kare, tab jo expression generate hoga p 1 - ∘ = 1 - χ1 p1 p 1∘ - p p 1∘ [ = χ 2...................(22) since χ 1 + χ 2 = 1 or 1 - χ 1 = χ 2 ] p solvent ∘ - p solution = χ solute p solvent ∘ ( ) Ab yahan p 1∘ - p pure solvent aur solution ke vapour pressure ka difference h jisko hum lowering in vapour pressure se represent karte h. Ab isko agar hum vapour pressure of pure solvent se divide kar de tab isko hum relative lowering in vapour pressure bolte h. Iss concept ke basis par hum Raoult’s Law ko redefine kar sakte h. The relative lowering in vapour pressure of an ideal solution containing the non-volatile solute is equal to the mole fraction of the solute at a given temperature Raoult’s Law as a special case of Henry’s Law Raoult’s law ke according ek solution meh volatile component vapour pressure ka expression hoga p 1 = p 1∘ χ 1 Yahan p 1∘ pure component ka vapour pressure aur p 1 solution meh jo component present h jiska mole fraction χ 1 Similarly agar hum gas in liquid ka solution consider kare toh isme gaseous component volatile hota h aur iske liye hum Henry’s law use karte h. p = K H. χ Yahan p pressure above the solution aur χ mole fraction ko represent kar raha h aur iss expression meh K H Henry’s law constant ko represent kar raha h. Agar hum dono expression ko consider kare toh dono expression meh partial pressure of volatile component or gas, solution ke mole fraction ke directly proportional h aur difference sirf proportionality constant ka aata h p 1∘ (Raoult’s Law) aur K H (Henry’s Law). So accordingly hum kah sakte h ki jab K H = p 1∘ hoga tab Raoult’s law- Henry’s law ka special case hota h. SOLVED EXAMPLES Question : At the same temperature, hydrogen is more soluble in water than helium. Which of them have a higher value of K H and why? Answer : As p A = K Hx A. Thus, at constant temperature, for the same partial pressure of different gases, x A = 1 / K H. In other words, solubility is inversely proportional to Henry's constant of the gas. Higher the value of K H, lower is the solubility of the gas. As H 2 is more soluble than helium, H 2 will have lower value K H than that of helium. Question : The partial pressure of ethane over a saturated solution containing 6.56 × 10 - 2g of ethane is 1 bar. If the solution contains 5.00 × 10 - 2 g of ethane, then what shall be the partial pressure of the gas ? Answer : Applying the relationship m = K H × p In the first case, 6.56 × 10 - 2g = K H × 1 bar or K H = 6.56 × 10 - 2g bar - 1 In the second case, 5.00 × 10 - 2g ( ) 5.00 × 10 - 2g = 6.56 × 10 - 2 g bar - 1 × p or p = 6.56 × 10 - 2 g bar - 1 = 0.762 b Vapour Pressure of Solutions of Solids in Liquids : Another important class of solutions consists of solids dissolved in liquid, for example, sodium chloride, glucose, urea and cane sugar in water and iodine and sulphur dissolved in carbon disulphide, but when we add any non - volatile solute there is a decrease in vapour pressure of solution and the vapour pressure of a solution at a particular temperature is found to be lower than that of pure solvent. This is because in the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced. Consequently, the number of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapour pressure is also reduced. The decrease in vapour pressure depends upon the quantity of non-volatile solute and its nature. Fig : Decrease in the vapour pressure of the solvent on account of the presence of solute in the solvent (a) evaporation of the molecules of the solvent from its surface is denoted by blue balls , (b) in a solution, solute particles have been denoted by green balls and they also occupy part of the surface area. According to Raoult's Law if p 1 is vapour pressure of the solvent, χ 1 be its mole fraction then, p1 ∝ χ1 p1 = χ1 p ∘ 1 The proportionality constant is equal to the vapour pressure of pure solvent, p ∘ 1. Fig : A plot between the vapour pressure and the mole fraction of the solvent is linear. Similarities between Henry's law and Raoult's Law : (a) Both Laws are applicable to volatile component in solution. (b). According to both the vapour pressure of a component is proportional to the mole fraction of that component. Differences : According to Raoult's Law the vapour pressure of the pure component is the proportionality constant whereas in case of Henry's Law an experimentally determined value gives the proportionality constant. Hello Points! Raoult’s Law as a special case of Henry’s Law : If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant K H differs from p 1∘ Thus, Raoult’s law becomes a special case of Henry’s law in which K H becomes equal to p 1∘ Ideal and Non-ideal Solutions : Liquid-liquid binary solution ko hum 2 types meh divide kar sakte h : (i) Ideal solutions (ii) Non-ideal solutions (i) Ideal solutions : Jo solutions over the entire range of concentration meh Raoult’s law follow karte h wo ideal solutions hote h. Ye solution generally tab bante h jab hum 2 components ko mix karte h jinka molecular size, structure, intermolecular forces identical hoti h. Iss solution meh dono components ki aapas meh intermolecular interaction (A-B attractions) in comparison to pure state meh intermolecular interaction (A-A and B-B attractions) ki magnitude same hoti h. Agar hum Raoult’s law ke basis par dono components ke partial pressure ko consider kare toh p 1 = p 1∘ χ 1 and p 2 = p 2∘ χ 2 Similarly, Dalton law of partial pressure ke basis par total pressure hoga p = p 1 + p 2 = p 1∘ χ 1 + p 2∘ χ 2 Ab Ideal solutions ke characteristics kya hote h inko discuss karte h (a) Heat change on mixing is zero : Ab kyunki solution meh jo dono component present unke mix hone par unke beech ki attractive forces ki magnitude meh koi change nahi aa raha isliye heat change ki value on mixing ΔH mixing inn solution ki zero hoti h. (b) Volume change on mixing is zero : Similarly agar hum volume of mixing ΔV mixing consider kare toh change in volume mixing ke baad ki value b zero hogi. For example agar hum 100 mL benzene ko 100 mL toluene meh add kare tab total volume hume 200 mL hi milti iska matlab jab humne inn dono solution ko mix kia tab volume meh koi difference nahi aaya. For example, ideal solutions benzene and toluene, n-hexane and n- heptane, chlorobenzene and bromobenzene. (ii) Non-ideal solutions : Jo solutions over the entire range of concentration meh Raoult’s law obey nahi karte karte wo non-ideal solutions hote h aur inn solutions ke liye p 1 ≠ p 1∘ χ 1 and p 2 ≠ p 2∘ χ 2 hota h. So inn solution ka vapour pressure Raoult’s law ke comparison higher or lower dono ho sakte h aur inn solutions ki ΔH mixing aur ΔV mixing ≠ 0 hoti h. Hello Points! Summary for ideal solutions : (i) It must obey Raoult’s law (ii) ΔH mixing = 0 (iii) ΔV mixing = 0 Summary for non-ideal solutions : (i) None of the components obey Raoult’s law (ii) ΔH mixing ≠ 0 (iii) ΔV mixing ≠ 0 Types of Non-Ideal solutions & Azeotropes or Constant Boiling Mixtures : Ab non-ideal solution, ideal solution se positive aur negative deviation show kar sakta h. (i) Non-ideal solutions showing positive deviation from Raoult’s law : Isko explain karne ke liye agar hum ek binary solution consider kare toh jab solution meh A-B interaction A-A aur B-B interaction se weaker hogi tab non-ideal solution positive deviation from Raoult’s law show karta h aur tab iss solution meh A aur B component ki evaporation tendency as compared to pure component jyada hoti h isliye dono component ke partial vapour pressure ki value Raoult’s law ke partial vapour pressure ke comparison jyada hoti h. Similarly total vapour pressure ki value b ideal solution ke total vapour pressure ke comparison jyada hogi. Isko agar mathematically dekhe p 1 > p 1∘ χ 1 and p 2 > p 2∘ χ 2..........(23) Accordingly total pressure ka expression hoga p > p1 + p2 p > p 1∘ χ 1 + p 2∘ χ 2..........(24) Figure : The vapour pressure of two component system as a function of composition showing positive deviation from Raoult’s law (dotted line represent ideal behaviour and solid line represent positive deviation) For example (i) ethyl alcohol and acetone (ii) Acetone and carbon disulphide (iii) benzene and acetone Explanation for positive deviation : Agar hum ethyl alcohol aur acetone consider kare toh isme ethyl alcohol molecule meh hydrogen bonding present hoti h. Lekin jab hum ethyl alcohol meh acetone daalte h, tab acetone ke molecules jo ethyl alcohol meh hydrogen present hote h usko weak kar deta h. Iss wajah se alcohol aur acteone ki escaping tendency increase ho jati hai aur issi wajah se solution ka vapour pressure Raoult’s law ke vapour pressure ke comparison jyada hota h. Similarly agar hum acetone aur carbon disulphide consider kare toh isme solute-solvent interactions in comparison to solute-solute aur solvent-solvent interactions weak hoti h isliye ye b positive deviation show karta h. Since non-ideal solution ki ΔH mixing ≠ 0 aur ΔV mixing ≠ 0 hoti h aur agar hum non-ideal solution showing positive deviation consider kare toh isme (a) ΔH mixing = + ve hoti h kyunki iss case meh A-A aur B-B bond ko break karne ke liye energy chahiye hoti h aur iss wajah se iss solution ka dissolution process endothermic hota h. Jis wajah se temperature increase karne par iss solution ki solubility increase hoti h. (b) Ab kyunki iss solution meh intermolecular forces ki magnitude decrease hoti h, matlab, iss solutions meh molecules loosely packed hote h jis wajah se iss solution ki ΔV mixing = + ve hoti h. (ii) Non-ideal solutions showing negative deviations from Raoult’s law : Isko explain karne ke liye agar hum ek binary solution consider kare toh jab solution meh A-B interaction A-A aur B-B interaction se stronger hoti h tab non-ideal solution negative deviation from Raoult’s law show karta h aur tab iss solution meh A aur B component ki evaporation tendency as compared to pure component kam hoti h isliye dono component ke partial vapour pressure ki value Raoult’s law ke partial vapour pressure ke comparison kam hoti h. Similarly total vapour pressure ki value b ideal solution ke total vapour pressure ke comparison kam hogi. Isko agar mathematically dekhe p 1 < p 1∘ χ 1 and p 2 < p 2∘ χ 2..........(25) Accordingly total pressure ka expression hoga p < p1 + p2 p < p 1∘ χ 1 + p 2∘ χ 2..........(26) Figure : The vapour pressure of two component system as a function of composition showing negative deviation from Raoult’s law (dotted line represent ideal behaviour and solid line represent negative deviation) For example : (i) Acetone and chloroform (ii) chloroform and diethyl ether (iii) phenol and aniline Explanation for negative deviation : Iske liye agar hum acetone aur chloroform consider kare toh jab hum inn dono solutions ko mixed karte h tab new attractive forces due to intermolecular hydrogen bonding generate hoti h jis wajah se iss solution ki attractive forces stronger ho jati h aur iss wajah se dono liquids ki escaping tendency decrease ho jati h aur issi wajah se solution ka vapour pressure Raoult’s law ke vapour pressure ke comparison kam hota h. Similarly phenol aur aniline meh b phenol ka proton aur aniline ke nitrogen par jo lone pair h wo yahan hydrogen bonding banate h jo similar molecules ki hydrogen bonding ke comparison strong hoti h iss wajah se ye mixture b negative deviation show karta h. Ab iss case meh b ΔH mixing ≠ 0 aur ΔV mixing ≠ 0 hoti h aur agar hum non-ideal solution showing negative deviation consider kare toh isme (i) ΔH mixing = - ve hoti h kyunki iss case attractive force increase hone par energy release hoti h aur iss wajah se iss solution ka dissolution process exothermic hota h aur further agar hum iss solution ko heat karte h tab iss solution ki solubility decrease hoti h. (ii) Ab kyunki iss solution meh intermolecular forces ki magnitude increase hoti h, matlab, iss solutions meh molecules tightly packed hote h jis wajah se iss solution ki ΔV mixing = - ve hoti h. Azeotropes or Constant Boiling Mixtures Ab kuch aise liquid mixture b hote h jisko jab hum boil karte h tab wo jo vapour generate karte h uski composition liquid ke equal hi hoti h aur ye constant temperature par boil hota h. Jis wajah se inn solutions ka jab hum distillation karte h toh ye aise distilled hote h jaise ye pure liquids h. Ab issi wajah se inn solution ko hum fractional distillation se b separate nahi kar sakte. Azeotropic mixture b 2 types ke hote h depending on ye Raoult’s law se kaise deviation show karta h. (i) Minimum boiling (ii) Maximum boiling azeotropes (i) Minimum boiling azeotropes : Jo solution large positive deviation from Raoult’s law show karte h wo minimum boiling azeotrope at a specific composition banate h. For example ethanol- water mixture meh water jyada hota h. Ab iss mixture ka agar hum fractional distillation kare toh isse hume 95% ethanol ki volume milti h. Lekin jab dono liquids ki composition equal hogi, tab dono liquid ethanol aur water ko hum distillation se separate nahi kar sakte. (ii) Maximum boiling azeotropes : Jo solution large negative deviation from Raoult’s law show karte h wo maximum boiling azeotropes at a specific composition banate h. For example nitric ( ) acid HNO 3 and water, iss azeotropic mixture meh 68 % nitric acid and 32 % water hota h jiska boiling point 393.5 K hota h. Ideal and Non-ideal (positive and negative deviations) & Azeotropic Mixtures Liquid-liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law : A. Ideal Solution : Solution which follow Raoult's Law over the entire range of concentration are ideal solutions.For such solution - ΔH mix = 0, ΔV mix = 0, ΔS mix > 0, ΔG mix < 0 (i.e. Spontaneous) At molecular level, ideal behaviour of the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present. For such solutions the observed vapour pressure is same as the vapour pressure calculated using Raoult's Law Vapour pressure is directly proportional to the tendency of evaporation, which is directly proportional to the tendency to leave liquid surface by molecules and tendency to leave is inversely proportional to the intermolecular force of attraction. If E A - A = E B - B = E A - B E A - A, E B - B, EA - B is the intermolecular force between respective molecules, responsible for attractive forces. Then vapour pressure (observed) is same as vapour pressure (calculated). In case of ideal solution the attractive forces are equal when solute and solvent are similar in structure and molar masses that is why benzene and toluene, n- hexane and n- Heptane are examples of ideal solutions. B. Non-ideal Solutions : When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution. Thus, mathematically we can say that when p sol ( )obs = (psol )cal The vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law.If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits negative deviation from Raoult’s law. Negative Deviation - Negative deviation arises when intermolecular forces of attraction between solute and solvent is more than intermolecular force of attraction in individual solute and solvent, that is E A - A and E B - B < E A - B. ΔH sol < 0, ΔV sol < 0, ΔS sol > 0, ΔG sol < 0 Fig : Negative deviation solutions can be represented graphically For Example : 1. In case of Phenol and Aniline the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding present in Aniline and Phenol. 2. Chloroform and Acetone : The chloroform molecule forms hydrogen bond with acetone molecule, which is stronger than the dipole-dipole bonding present in Chloroform and Acetone Positive deviation - Those type of solutions in which the calculated vapour pressure is lesser than that of observed vapour pressure. In case of positive deviation the intermolecular forces of attraction between solute and solvent is less than cohesive intermolecular force of attraction in solute and solvent, that is E A - A and E B - B > E A - B ΔH sol > 0, ΔV sol > 0, ΔS sol > 0, ΔG sol < 0 For Example 1. Ethanol and Acetone : Pure ethanol, molecules are hydrogen bonded when acetone is added it gets between the molecules of ethanol and breaks the hydrogen bond present in the molecule and the interactions get weakened. 2. Water and Ethanol water has strong hydrogen bonding and ethanol has weak hydrogen bonding so when ethanol is mixed with water it breaks the strong hydrogen bonds and positive deviation is shown. Fig : Positive deviation solutions can be represented graphically Azeotropy Some liquids on mixing, form azeotropes which are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation because it will get distilled but the composition will remain the same. There are two types of azeotropes called Minimum boiling azeotrope and Maximum boiling azeotrope shown by the non- ideal solutions. Maximum Boiling Azeotrope - In case of a solution showing negative deviation the total vapour pressure is less than expected by Raoult's law and when such solutions are heated their boiling points are increased. One of the intermediate composition of the solution boils at constant temperature, the total vapour pressure will be least and boiling point will be maximum. This kind of Azeotropes are known as Maximum Boiling Azeotropes as they have a composition that has a Maximum Boiling point. For example - 68% Nitric Acid + 32% water, in this case the boiling point of the solution(azeotrope) is greater than the compounds separately. Minimum Boiling Azeotrope - The solutions which show positive deviations when heated achieve an intermediate composition when the pressure will be highest and temperature will be minimum, the composition of the liquid and vapour phase is same and it will be boiling at a constant temperature such solutions are called minimum boiling azeotropes because they have constant minimum boiling point. For example - 95.5% Ethanol + 45% Water, in this the boiling point of solution(azeotrope) is lesser than that of pure compounds. Colligative Properties and Determination of Molecular Mass : Aise dilute solution jisme non-volatile solutes present h wo kuch characteristics properties show karte h jo nature of solute par depend na kar number of solute particles par depend karta h matlab ye molar concentration of solute par depend karta h. Inn properties ko hum colligative properties bolte h. The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties The four important colligative properties are : (i) Relative lowering in vapour pressure. (ii) Elevation in boiling point (iii) Depression in freezing point (iv) Osmotic pressure (i) Relative lowering in vapour pressure : Jaise humne pehle pada tha ki jab hum non-volatile solute ko solvent meh add karte h tab solution ka vapour pressure decrease hota h. Suppose χ 1 mole fraction of solvent h, χ 2 mole fraction of solute h. Similarly p 1∘ pure solvent ka vapour pressure h aur p solution ka vapour pressure h. Ab kyunki solute non-volatile h isliye solute ka vapour pressure meh koi contribution nahi hota, so solution ka overall vapour pressure solvent ki taraf se lgta h matlab solution aur solvent ka vapour pressure equal honge. p = p1 Lekin agar hum Raoult’s law consider kare, toh isme solvent ka vapour pressure product of vapour pressure in pure state aur mole fraction ke equal hota h. p 1 = p 1 χ 1or p = p 1 = p 1 χ 1..........(27) ∘ ∘ Ab kyunki mole fraction χ 1 hmesha less than 1 hoga iss wajah se solution ka vapour pressure b p 1 se kam hoga (pure solvent). Iss ∘ basis par agar hum lowering in vapour pressure consider kare : Δp 1 = p 1∘ - p 1 Δp 1 = p 1 - p 1 χ 1 ∘ ∘ ( Δp 1 = p 1 1 - χ 1 ∘ ) Lekin 1 - χ 1 = χ 2 jiske according Δp 1 = p 1 χ 2 ∘ Δp 1 = χ2 p1 ∘ p 1 - p1 ∘ Δp 1 = = χ 2..........(28) p1 p1 ∘ ∘ Yahan (p ∘ 1 - p1 ) vapour pressure of pure solvent aur solution ka p 1 - p1 ∘ difference h aur relative lowering in vapour pressure ko p1 ∘ represent kar raha h. Thus the relative lowering in vapour pressure of an ideal solution containing the non-volatile solute is equal to the mole fraction of the solute at a given temperature Determination of molar mass of a solute from relative lowering in vapour pressure So according to Raoult’s law, relative lowering in vapour pressure ka expression hota h. Δp 1 p 1∘ - p 1 = = χ2 p1 p1 ∘ ∘ Agar hum w 1 aur w 2 ko solvent aur solute ka weight aur M 1 aur M 2 ko inka molar mass consider kare, so mole fraction ke according n2 Mole fraction of solute χ 1 = n1 + n2 Jahan n 1 aur n 2 solvent aur solute ke moles h w1 w2 n1 = ,n = M1 2 M2  w2 / M2 ∴ χ2 = w2 / M2 + w1 / M1 So relative lowering in vapour pressure ke expression ko hum likh sakte h p 1 - p1 ∘ w2 / M2 =..........(29) p 1∘ w2 / M2 + w1 / M1 Lekin dilute solution ke case meh n 2 in comparison n 1 small hota h isliye n 2 ko hum denominator meh neglect kar sakte tab jo expression banega n 1 + n 2 ≈ n 1 ( ) p 1 - p1 ∘ w2 / M2 = p 1∘ w1 / M1 p 1 - p1 ∘ w2 × M1 =..........(30) p 1∘ w1 × M2 w2 × M1 or M 2 = ∘..........(31) p1 - p1 w1 × ∘ p1 So aise hum relative lowering use kar molar mass of solute calculate kar sakte h. SOLVED EXAMPLES Question : The vapour pressure of pure benzene at a certain temperature is 0.650 bar. A non-volatile, non-electrolyte solid weighing 0.5 g is added to 39.0 g of benzene (molar mass 78gmol - 1). The vapour pressure of the solution then is 0.645 bar. What is the molar mass of the solid substance ? Answer : Here, we are given that w 2 = 0.5g, w 1 = 39.0g, M 1 = 78gmol - 1 p ∘ = 0.650 bar, p s = 0.645 bar Substituting these values in the formula, p ∘ - ps n2 w2 / M2 w 2M 1 = = = , we get p∘ n1 w1 / M1 w 1M 2 0.650bar - 0.645bar 0.5g × 78gmol - 1 = = 0.650bar 39.0g × M 2 0.5 × 78 0.650 or M 2 = × gmol - 1 = 130gmol - 1 39.0 0.005 (ii) Elevation in boiling point : Isko padhne se pehle hume ye pta hona chahiye ki boiling point kya hota h? So boiling point wo temperature hota jis par liquid ka vapour pressure atmospheric pressure ke equal hota h aur iss temperature par liquid and vapour state equilibrium meh hote h. Boiling point of a liquid is the temperature at which its vapour pressure become equal to the atmospheric pressure Jaise humne pehle pada jis solution meh non-volatile solute hota h uska vapour pressure solvent ke vapour pressure se kam hota h aur jab hum iss solution ko high temperature par heat karte h tab iska vapour pressure atmospheric pressure ke equal ho jata h. Jis wajah se solution ka boiling point pure solvent ke comparison hmesha jyada hota h. For example water ka vapour pressure 373 K par 1.013 bar ( or 1 atm) hota h matlab 373K par water boil karta h kyunki iss temperature par iska vapour pressure atmospheric pressure ke equal hota h. Lekin agar hum sucrose consider kare toh iska vapour pressure 373K par 1.013 bar se kam hota h. Jis wajah se iska solution 373 K par boil nahi karta. Lekin jab hum yahan temperature increase karenge tab iska vapour pressure 1.013 bar ke equal ho jayega. Jis basis par hum yahan conclude kar sakte h ki solution ka boiling hmesha pure solvent solution ke boiling point se jyada hoga. Isko hum graphically b dekh sakte h. Figure : The vapour pressure curve for solution lies below the curve for pure water. The diagram shows that ΔT b denotes the elevation of boiling point of a solvent in solution. Ab hum graph ke through kah sakte h ki jis temperature par vapour pressure of solvent atmospheric pressure ke equal hoga uss temperature ko T b se represent karte h aur jis temperature par ∘ solution ka vapour pressure atmospheric pressure ke equal hoga usko hum T b se represent karte h aur inn dono ke difference ko hum ΔT b se represent karte h. Ab Elevation in boiling point meh T b greater than T b∘ hota h so isko hum mathematically consider kare ΔT b = T b - T b ∘ Iske sath Elevation in boiling point (ΔTb ) molal concentration of solution ke directly proportional hota h. ΔT b ∝ m ΔT b = K bm..........(32) Iss expression meh m molality of solution h, aur K b molal boiling point elevation constant or molal boiling point constant or ebullioscopic constant h.The unit of K b is K kg mol - 1. Agar hum molal boiling point elevation constant, K b ko define kare The elevation in boiling point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 gram of the solvent Thus, elevation in boiling point is directly proportional to molal concentration of the solute (number of molecules) Determination of molar mass of solute from elevation in boiling point temperature Agar hum w 2 g non-volatile solute ka weight, w 1 g solvent ka weight aur M 2 molar mass of solute consider kare. So according to molality, m moles of solute × 1000 m= wt. of solvent in grams  w2 Yahan moles of solute = honge jis basis par molality hogi M2 w 2 × 1000 m= M2 × w1 Ab isko elevation in boiling point expression meh substitute kare w 2 × 1000 ΔT b = K b ×..........(33) M2 × w1 Aur iske through hum molar mass calculate kar sakte h jiska expression w 2 × 1000 M2 = Kb × ΔT b × w 1 SOLVED EXAMPLES Question : 36 g of glucose, C 6H 12O 6 is dissolved in 1 kg of water in a saucepan. At what temperature will the solution boil at 1.013 bar pressure ? K b for water is 0.52 K kg mol - 1. Answer : Here, we are given w 2 = 36g, w 1 = 1kg = 1000g, K b = 0.52kgmol - 1 ) M 2 (for glucose, C 6H 12O 6 = 72 + 12 + 96 = 180gmol - 1 1000K bw 2 1000 × 0.52Kkgmol - 1 × 36g ΔT b = = = 0.104K w 1M 2 1000g × 180gmol - 1 As water boils at 373.15 K at 1.013 bar pressure, therefore, boiling point of solution = 373.15 + 0.104K = 373.254K. (iii) Depression in freezing point Isko padhne ke liye hume pehle ye pta hona chaiye ki freezing point kya hota h, so freezing point wo temperature hota h, jispar kisi substance ki solid aur liquid state ka same vapour pressure hota h aur iss temperature par solid and liquid state equilibrium meh hote h. Ab iss case meh jab hum ek non-volatile solute ko solvent meh add karte h tab uss solution ka freezing point pure solvent ke comparison hmesha kam hota h. Isko hum graphically dekhe toh Figure : The graph showing T f depression of the freezing point of a solvent in a solution Ab hum graph ke through kah sakte h jis temperature par solid aur liquid solvent equilibrium meh hote h usko hum T f se represent ∘ karte h aur jis temperature par solid solvent and liquid solution sath meh honge usko hum T f se represent karte h. Ab kyunki solution ka freezing point (T f) pure solvent ke comparison kam hota h isko agar hum mathematically consider kare. ΔT f = T f∘ - T f ( ) Iske sath Depression in freezing point ΔT f molal concentration of solution ke directly proportional hota h. ΔT f ∝ m ΔT f = K fm..........(34) Iss expression meh m molality of solution h, aur K f molal freezing point depression constant or molal cryoscopic constant h.The unit of K f is K kg mol - 1. Agar hum molal freezing point depression constant, K f ko define kare The depression in freezing point for 1 molal solution i.e., a solution containing 1 gram mole of solute dissolved in 1000 g of solvent. The depression in freezing point temperature is directly proportional to the molal concentration of the solute Determination of molar mass of solute from Depression in freezing point temperature Agar hum weight of solute w 2 g, weight of solvent w 1 consider kare aur molar mass of solute M 2 consider kare. Tab humare pass molality ka expression hoga.  w 2 × 1000 m= M2 × w1 Ab m ki value hum depression in freezing point expression meh put kare tab humare pass expression hoga K f × w 2 × 1000 ΔT f = w1 × M2 K f × w 2 × 1000 M2 =..........(35) w 1 × ΔT f Aise hum depression in freezing point ke basis par molar mass of the solute calculate kar sakte h. Ab hum K f aur K b ko ΔH fusion aur ΔH vapourisation se b relate kar sakte h jiske according humare pass 2 expression hote h. R × M × T 2f Kf =..........(36) 1000 × ΔH fus R × M × T 2b Kb =..........(37) 1000 × ΔH vap SOLVED EXAMPLES Question : Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g mol - 1) in 250 g of water. (K f of water = 1.86Kkgmol - 1) Answer : ΔT f = K f × m  Kf × WB Tf - Ts = ∘ M B × W A(kg) T f∘ for water = 0 ∘ C = 273K ( ) M B C 6H 12O 6 = 12 × 6 + 12 × 1 + 6 × 16 = 180g / mol 1.86 × 60 × 1000 273 - T s = 180 × 250 273 - T s = 2.48 T s = 273 - 2.48 T s = 270.52K T f = freezing point of solvent ∘ T s = freezing point of solution W B = given mass of solute M B = molar mass of solute W A = mass of solvent K f = molal depression constant (iv) Osmosis and Osmotic Pressure Ab isko padhne ke liye hume ye pata hona chaiye ki osmosis kya hota h? So iske liye hum ek sucrose ka aqueous solution inverted thistle funnel meh consider karte h jiske niche semi permeable membrane lagi hui h (semi permeable membrane jo animal bladder or parchment paper se bani hui hoti h) ab thistle funnel ko ek beaker meh rakhte h jisme water present h. Ab iss semi permeable membrane se sirf solvent molecules hi pass karte h, solute nahi karte. So basically iske through hum kah sakte h ki ab water molecules pure solvent se solution meh move kate h jis wajah se thistle funnel meh solution ka level increase karne lagta h iss process ko hi hum osmosis kahte h. The phenomenon of the flow of solvent through a semipermeable membrane from solvent to the solution is called osmosis. Osmosis solutions ki different concentration par b ho sakta h. Iss case meh, solvent molecules less concentrated solution se more concentrated solution through semipermeable membrane move karta h. Nature meh semipermeable membrane plant or animal dono mein hoti h for specific functions. Example pig’s bladder or parchment lekin ye laboratory ke liye useful nahi hota inke imperfect nature ki wajah se. Agar ek artificial membrane ka example consider kare - [ ] film of gelatinous precipitate of cupric ferrocyanide Cu 2 Fe(CN) 6 ye membrane dikhne meh toh continuous sheets or films ki tarah hoti h lekin isme network of submicroscopic holes or pores b hote h. Isme solvent molecules jaise water pass kar sakta hai lekin solute particles isme se pass nahi ho sakte aur iss type ki membrane ko hum semi permeable membrane (SPM) bolte h. Consider an activity to understand Osmosis Sabse pehle hum 2 eggs lete h jiske outer hard shell ko hum dil. HCl solution meh dissolve kar remove karte h. Ab isme se ek egg ko hum distilled water aur dusre egg ko hum sodium chloride ke saturated solution meh dipped karte h. Kuch minutes ke baad jo egg distilled water meh tha wo phul (swell) jata h aur jo egg saturated sodium chloride solution meh tha wo sikud (shrink) ho jata h. Ab iss observation ko hum osmosis ke through samjh sakte h jiske according egg ki skin semipermeable membrane ki tarah kaam kar rahi h. First case meh egg ke bahar water(solvent) ki concentration jyada h iss wajah se water egg meh entry karta h aur egg ko swell karta h. Similarly dusre case meh water ki concentration egg meh jyada h isliye water(solvent) egg se bahar niklata h aur issi wajah se egg shrink ho jata h. Osmotic Pressure : Jaise humne pada ki jab humne sugar ka aqueous solution inverted thistle funnel meh lia tha, tab semipermeable membrane ki wajah se water molecules osmosis ke through sugar solution meh enter karte h. Ab jaise funnel meh solution ka level increase hoga waise hi solution par additional hydrostatic pressure lagta h. Iss pressure ki wajah se solvent ka funnel meh inflow kam ho jata h ya fir hum ye kah sakte h ki ye pressure funnel meh solvent ke inflow ko oppose karta h. Lekin ek particular height tak solution ka level increase hota h aur jis stage par hume equilibrium milti h uss stage par liquid ka hydrostatic pressure liquid ko semipermeable membrane ke through thistle funnel ke andar jane ki tendency ko balance kar deta h matlab iss stage par osmosis process nahi ho raha hota. The equilibrium hydrostatic pressure on the solution due to osmosis of the pure solvent into it is a measure of osmotic pressure. Isko samjhne ke liye agar hum ek apparatus consider kare jisme 2 vessels h jo ek semi permeable membrane ke through connected h aur ye dono hi water-tight frictionless pistons se fit kiye gye h. Ab ek compartment meh solution consider karte h aur ek compartment meh solvent consider karte h. Ab isme osmosis ki wajah se solvent molecule through semipermeable membrane solution meh flow karte h. Jis wajah se solution ka piston bahar nikalne lagta lekin agar hum solution side par extra pressure lagyein tab solution ka piston bahar nahi niklega matlab iss pressure ki wajah se osmosis ruk jata h isliye iss pressure ko hum osmotic pressure bolte h. Figure : The excess pressure equal to the osmotic pressure must be applied on the solution side to prevent osmosis. Thus, osmotic pressure is defined as : The minimum excess pressure which must be applied to a solution to prevent the passage of solvent into it through a semipermeable membrane. It is denoted by π. Lekin jab kabhi bhi solution par pressure osmotic pressure se jyada ho uss case meh solvent ka flow solution se through semi permeable membrane pure solvent ki taraf hota h. Osmotic pressure -colligative property Scientist Vant’s Hoff’s ke according dilute solutions ke liye, osmotic pressure ko hum iss expression ke through define kar sakte h. π = cRT Yahan c solution ki molar concentration h, T temperature aur R gas constant h. So basically hum kah sakte h π ∝ c kyunki T aur R constant h. Determination of molar mass from osmotic pressure Vant’s Hoff Equation ke according π = cRT n Lekin c = hota h aur yahan n number of moles of solute dissolved V in V litres of volume of solution hota h n ∴ π = RT v πV = nRT Iss equation ko hum vant’s hoff equation for dilute solutions bolte h. ( ) Jisme number of moles of solute ko hum mass w 2 and its molar ( ) mass M 2 se calculate kar sakte h. w2 n= M2 Iski value equation meh put kare w 2RT w 2RT π= or M 2 = VM 2 Vπ Inn equations se hum molar mass osmotic pressure ko use kar calculate kar sakte h. Biological significance of osmosis Plants jab soil se roots ke through water absorb karte h tab ye osmosis ke through water ko different parts of plants meh pass karte h. Plants ke cells meh fluids hoti h jisko hum cell sap kehte h aur cells ki jo walls hoti h wo living cytoplasmic membrane se bani hui hoti h jo ek semi permeable membrane ki tarah kaam karti h. Agar ye cells water or dilute solution ke contact meh aate h jiska osmotic pressure cell sap ke pressure se kam h tab water ki itni tendency hoti h ki wo cell wall ke through cell meh enter kar sakta h, jis wajah se cell toot (rupture) jata h ya fir swelling show karta h aur iss process ko hum hemolysis kahte h. Lekin jab ye cells uss solution jiska osmotic pressure cell sap ke pressure se jyada h uske contact meh aata h tab water ki itni tendency hoti h ki wo cell se bahar nikal jata h aur cell ko shrink kar deta h aur iss process ko hum plasmolysis kehte h. Osmosis ke through hum salt and sugar ko pickles aur jams ki preservation ki form meh b use kar sakte h. Isotonic Solutions Jaise ki humne abhi tak pada ki different solutions ke different vapour pressure hote h jis wajah se inke osmotic pressure b different honge. Ab jab aise 2 solutions jo semipermeable membrane se separated h aur isme solvent molecules ka flow lower osmotic pressure solution to higher osmotic pressure solution ki taraf ho raha ho aur ye tab tak ho jab tak dono solution ka osmotic pressure equal ho aur iss stage par osmosis process nahi ho raha hota. The solution having same osmotic pressure at the same temperature are called isotonic solution or isosmotic solutions From the equations π = cRT, jab do solutions ki concentration same hogi tab unn dono solutions ka osmotic pressure same temperature par same hoga. So iss basis par hum isotonic solution ko iss form meh b define kar sakte h. Solutions of equimolar concentrations at the same temperature have same osmotic pressure are called isotonic solutions Hypertonic Solutions Agar ek solution ka osmotic pressure dusre solution ke osmotic pressure se jyada h tab wo solution hypertonic solution hote h. For example 0.1 M urea solution ka osmotic pressure 0.05 M sucrose solution se jyada hota h. Iss wajah se 0.1 M urea solution hypertonic solution h. Hypotonic Solutions similarly agar ek solution ka osmotic pressure dusre solution ke osmotic pressure se kam h tab wo solution hypotonic solution hote h. For example 0.05 M sucrose solution ka osmotic pressure 0.1 M urea solution se kam hota h. Iss wajah se 0.05 M sucrose solution hypotonic solution h. Inn dono solution ke basis par hum kah sakte h ki hypertonic solution dusre solution se jyada concentrated hota h aur hypotonic solution dusre solution se less concentrated hota h. SOLVED EXAMPLES Question : 500cm 3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10 - 3 bar. Calculate the molar mass of the protein. Answer : Here, we are given : Mass of solute (protein), w 2 = 1.26g Volume of the solution (V) = 500cm 3 = 0.500L π = 2.57 × 10 - 3 bar, T = 300 K, R = 0.083 L bar K - 1mol - 1 w 2RT Substituting these values in the formula, M 2 = , we get πV 1.26g × 0.083LbarK - 1mol - 1 × 300K M2 = = 24415gmol - 1 2.57 × 10 - 3bar × 0.500L Question : A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate the osmotic pressure of the solution. Given solution constant (R ) = 0.082057 L atm mol - 1K - 1 Answer : [ K 4 Fe(CN) 6 ] ⇔ 4K + + [Fe(CN)6 ]4 - 1-α 4α α. Total = 1 + 4α i = 1 + 4α = 1 + 4 × 0.5 = 3 Apply π = iCRT 1 π=3× × 0.0821 × 300 = 7.389 Atm 10 Colligative Properties and Determination of Molecular Mass : The properties that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Such properties are called Colligative properties There are four colligative properties: (1) Relative lowering of vapour pressure of the solvent (2) Depression of freezing point of the solvent (3) Elevation of boiling point of the solvent and (4) Osmotic pressure of the solution. A. Relative Lowering of Vapour Pressure - As we know that when a non- volatile solute is added to a solvent the vapour pressure of pure solvent decreases. According to Raoult's Law, p S = p A = p A∘ χ A p A = Vapour Pressure of pure solvent ∘ χ A = Mole fraction of solvent For a solution χ A + χ B = 1 or χ A = 1- χ B p S = p A∘ (1- χ B ) pS or = 1- χ B pA ∘ pS p A∘ - p S χB = 1 - = pA pA ∘ ∘ p A∘ - p S or = χB pA ∘ p A∘ - p S = lowering in Vapour pressure ( Not a colligative property) p A∘ - p S = Relative Lowering of vapour pressure p A∘ ( ) pA - pS ∘ nB nB = χB = pA ∘ nA + nB nA + nB Here n A and n B are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n A < < n B, hence neglecting n B in the denominator we have p A∘ - p S nB WB × MA = = pA ∘ nA M