Solutions Chemistry Questions PDF

Solution Question 1 Explain solutions with positive deviations from Raoult's law. Answer: dig from notes \(i) A solution or a liquid mixture which has higher vapour pressure than theoretically calculated by Raoult's law or higher than those of pure components is called a nonideal solution with p...

Solution Question 1 Explain solutions with positive deviations from Raoult's law. Answer: dig from notes \(i) A solution or a liquid mixture which has higher vapour pressure than theoretically calculated by Raoult's law or higher than those of pure components is called a nonideal solution with positive deviation. \(ii) In these solutions, solute-solvent intermolecular attractions are weaker than those between solvent-solvent and solute-solute interactions. \(iii) For example, solutions of acetone and ethanol, carbon disulphide and acetone, etc. Question 2 Explain solutions with negative deviations from Raoult's law. Answer: dig from notes \(1) A solution or a liquid mixture which has lower vapour pressure than theoretically calculated by Raoult's law or lower than those of pure components is called a nonideal solution with negative deviation \(2) In these solutions, the intermolecular interactions between solvent and solute molecules are stronger than solvent-solvent or solute-solute interactions. \(3) For example, solutions of phenol and aniline, chloroform and acetone, etc. Question 3 State and explain Raoult's law for solutions of nonvolatile solutes. Answer: \(a) Statement of Raoult's law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature. \(b) Explanation : Let P0 and P be the vapour pressures of a pure solvent and a solution respectively. If x1 is the mole fraction of the solvent then Raoult's law can be represented as, P = x1P0 For a binary solution containing one solute, if x1 and x2 are mole fractions of a solvent and a solute respectively then, x1 + x2 = 1 ∴ x1 = 1 -- x2 ∴ P = x1P0 = (1 -- x2) P0 = P0 -- x2P0 = P0 -- P = x2P0 ∴ P0 -- P = x2P0 ∴ x2 = P0−PP0 P0 -- P = ΔP is the lowering of vapour pressure ∴ x2 = ΔP/P0 In the equation, P0 -- P/P0 is called relative lowering of vapour pressure. Hence Raoult's law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute. Question 4 Explain the effect of temperature on the vapour pressure of a liquid. Answer: The vapour pressure of a liquid is the pressure of the vapour in equilibrium with the liquid at a given temperature. The evaporation of a liquid requires thermal energy. Hence, as temperature rises, the vapour pressure rises until it becomes equal to the external pressure, generally the atmospheric pressure, 101.3 kNm-2 (1 atm). This temperature is called the normal boiling point of the liquid. Question 5 What is abnormal colligative property? Explain the reasons. Answer: Abnormal colligative property : When the experimentally measured colligative property of a solution is different from that calculated theoretically by the van't Hoff equation or by the laws of osmosis, then the solution is said to have abnormal colligative property. Explanation : The colligative property depends on the number of solute particles in the solution but it is independent of their nature. Abnormal values of them arise when the dissolved solute undergoes a molecular change like dissociation or association in the solution. The observed colligative property (or abnormal colligative property ) may be higher or lower than the theoretical value. \(i) Dissociation of the solute molecules : When a solute like an electrolyte is dissolved in a polar solvent like water, it undergoes dissociation, which results in the increase in the number of particles in the solution. Hence, the observed value of the colligative property becomes higher than the theoretical value, e.g., when one mole of KCl is dissolved in the solution then due to dissociation, KCl → K+ + Cl--, the number of particles increases, hence, the colligative properties like osmotic pressure elevation in the boiling point, etc. increase. \(ii) Association of the solute molecules : When a solute like a nonelectrolyte is dissolved in a nonpolar solvent like benzene, it undergoes association forming molecules of higher molecular mass. Hence, the number of the particles in the solution decreases. Therefore the colligative properties like osmotic pressure, elevation in the boiling point, etc., are lower than the theoretical value, e.g., nA → An. 2CH3COOH → (CH3COOH)2 2C6H5COOH → (C6H5COOH)2 Question 6 Obtain a relation between degree of dissociation and molar mass for an electrolyte. Answer form notes Q7 explain vantoff factor and from notes Q8. Explain elevation in boiling point and depression in freezing point with suitable graph it\'s unit Ans from notes