Colligative Properties PDF

Chapter 1 – Colligative Properties Terms ▪ Atomic weight ▪ Molecular weight ▪ Mole (n) - Mole is the mass of the substance containing Avogadro’s number (6.023 X 1023) of particles. Solutions ▪ Solution – define ▪ Concentrated solution – define ▪ Dilute solution – define Comparison of concentrated...

Chapter 1 – Colligative Properties Terms ▪ Atomic weight ▪ Molecular weight ▪ Mole (n) - Mole is the mass of the substance containing Avogadro’s number (6.023 X 1023) of particles. Solutions ▪ Solution – define ▪ Concentrated solution – define ▪ Dilute solution – define Comparison of concentrated solution and dilute solution  Concentration Units of Solutions Expressions used to measure the concentrations of a solutions 1. Normality (N) 2. Molarity (M) 3. Molality (m) 4. Mole Fraction (χ) Explore and learn the following about the above mentioned concentration units Definition Expression Units Relation between normality and molarity Compare the concentration units  Concentration Units of Solutions 1. Normality (N) – number of gram equivalents of solute dissolved in a litre of solution. Equivalent weight – the mass of an element or compound or ion which combines or displaces 1 part of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine by mass. Basicity – number of hydrogen or hydronium ions produced by one molecule of acid when it dissolves in water. Acidity – number of hydroxyl or hydroxide ions produced per molecule of base when it dissolves in water. Total Charge on cation or anion Example – CaCl2 and Na2SO4. Ca 2+ 2Cl- 2Na+ SO4 2- Total positive charge on Ca 2+ = 2 Or Total negative charge on 2Cl- = 2  Question Find the equivalent weights of following: 1. Sulphuric acid 2. Acetic acid 3. Calcium 4. Sodium 5. Potassium sulphate Steps Write the formula of the compound or element. Find the molecular mass of compound or atomic weight if element. Calculate the equivalent weight using the formula Expression of normality (N)  Questions 1. What is the normality of solution that contains 1.5 g of calcium hydroxide dissolved in 850mL? 2. What is the normality of solution that contains 50 g of sulphuric acid dissolved in 15L?  Questions 1. What is the normality of solution that contains 1.5 g of calcium hydroxide dissolved in 850mL? Weight of solute [Ca (OH)2] = 1.5 g Volume of solution in mL = 850 mL 2. Molarity (M) – number of moles of solute dissolved per litre of solution at a particular temperature. Expressions of molarity (M) 1. What is the molarity of solution that contains 1.5 g of calcium hydroxide dissolved in 850 mL? Relation between normality and molarity Molarity x Molecular mass = Normality x Equivalent weight Question 1. What is the normality of 3M sodium hydroxide solution? 2. What is the normality of a solution containing 0.248 moles of sulphuric acid dissolved in 250mL of solution? 3. Molality (m) – number of moles of solute per kg of solvent or 1000 g of solvent. Expressions Questions 1. What is the molality of solution containing 18 g of urea in 1.5 Kg of solvent? Weight of urea = 18 g Weight of solvent = 1.5Kg = 1500g Molecular mass of urea, (NH2)2CO = ?  Questions 1. 4g of sugar is dissolved in 350 mL of water at 80°C. What is the molality of the sugar solution? Given: Density of water at 80°C is 0.975g/mL 2. How many grams of water must be used to dissolve 100g of sucrose to prepare 0.2 molal solution? A molar solution of a solute in water is more concentrated than molal solution. Explain Molar solution – one mole of solute in 1000mL of water. Molal solution – one mole of solute in 1000g of water. At room temperature, density of water is slightly less than 1. Therefore, the volume of solvent corresponding to 1000g of water would be greater than 1000 mL. 4. Mole Fraction (χ) – ratio of the number of moles of a given component to the total number of moles of all the components in the mixture. OR Example Solution containing component A & B are present in a mixture No. of moles of component A = nA No. of moles of component B = nB Sum of mole fractions of all the components in a solution is always one Mole fraction is a dimensionless quantity as it is a ratio of two numbers representing number of moles Questions 1) 0.1 mole of sodium chloride is dissolved in 100g of water. Find the mole fraction of solute and the solvent? 2) A solution is prepared by mixing 25g of water and 25g of ethanol. Calculate the mole fraction of water? Note: Concentration units which vary with temperature are molarity and normality. Reason – involve volumes of the solution and volume vary with temperature. Concentration unit which are independent of temperature are molality and mole fraction. Reason – involve weights of the solute and solvent which do not change with temperature. Vapour pressure of liquids – the pressure exerted by vapour in equilibrium with a liquid at a given temperature. In the closed container, there is the equilibrium established between the vapour and liquid phases when rate of evaporation is equal to the rate of condensation. Raoult’s law for solution containing non-volatile solute. The vapour pressure of solution containing non-volatile solute at a particular temperature is directly proportional to the mole fraction of the solvent in the solution. Or The vapour pressure of a solution containing non-volatile solute at a particular temperature is equal to the product of the vapour pressure of the solvent in the pure state and its mole fraction in the solution. Expression of Raoult’s Law When a solution contains non-volatile solute, there is no contribution of the non-volatile solute to the total vapour pressure of the solution. Therefore the vapour pressure of a solution containing non-volatile solute is actually the vapour pressure of the solvent in the solution. V.P of solution = V.P of solvent in solution.  Colligative properties The properties of the solution which depend upon the number of solute particles but not upon their nature of the solute. Important colligative properties 1. Relative lowering of vapour pressure (∆P) 2. Elevation of boiling point or Ebullioscopy (∆Tb) 3. Depression of freezing point or Cryoscopy (∆Tf) 4. Osmotic pressure (π) Note – solution of volatile solvent, non volatile solute and non-electrolyte (solute which do not dissociate or associate in solution) 1. Relative lowering of vapour pressure Vapour pressure of volatile solvent gets lowered when non-volatile solute is dissolved in it. The non-volatile solute has no contribution towards vapour pressure. Lowering of vapour pressure – difference between the vapour pressure of pure solvent and vapour pressure of solution. Why is there lowering of vapour pressure when non-volatile solute is added? In pure solvent the whole surface is occupied by solvent molecules and escaping tendency of solvent molecules is more. In solution the surface is occupied partly by molecules of solvent and partly by molecules of solute. There is decrease in the number of solvent molecules on the surface of solution, thus decreasing the escaping tendency of solvent molecules and lowering the vapour pressure of solution. Derivation of mathematical expression for relative lowering of vapour pressure. According to Raoult’s Law Substitute (i) in (ii) Rearrange Substitute (iv) in (iii) Relative lowering of vapour pressure is equal to mole fraction of non-volatile solute. Or * Relative lowering of V.P for a binary solution containing non-volatile solute and a volatile solvent is equal to the mole fraction of solute. Determination of molecular mass from Relative lowering of Vapour Pressure. Derivation of expression to calculate molecular mass from relative lowering of vapour pressure. Substitute value of χsolute from eq. (i) For dilute solution nsolute is very small compared to nsolvent so neglected. On rearranging the right hand side of equation wsolute = weight of solute Wsolvent = weight of solvent msolute = molecular mass of solute Msolvent = molecular mass of solvent Questions 1. If 0.34 moles of a non-volatile solute is dissolved in 3 moles of water. What is the vapour pressure of the resulting solution.? The vapour pressure of pure water is 23.8 torr at 25°C. 2. A non-volatile compound was used to prepare a solution. A solution contains 5g of non-volatile compound dissolved in 100g of water and has a vapour pressure of 754.5 mm Hg at a normal boiling point of water. Calculate the molecular mass of the compound?