Lec.11

1. Lec.11 Liquid State & Solution

2. Liquid State & Solution • Solution may be defined as a homogeneous mixture of two or more substances whose composition can be continuously varied within certain limits. • Binary Solutionare mixtures of only two chemical substances i.e. it has two constituents. The substance which present in a large excess of other is usually called the solvent and the substance present in lesser quantity is the solute. The solvent may be a gas, liquid, or solid and so may the solute. e.g.: • Liquid in liquid: a solution of little alcohol in a great deal of water, where alcohol is the solute, and water is the solvent. • Solid in liquid: a solution of sugar in water. • Gas in liquid: a solution of oxygen in water or coca (CO2in water) • Note: solutions properties e.g. density, surface tension or viscosity, depends on the kind of solute molecules present as well as their number in certain quantity of the particular solvent.

3. Methods of expressing composition 1-Molarity: moles of solute in one liter of solution Molar solution(1Molar) : one mole of solute in one liter of solution. 2- Molality: moles of solute in 1000g of solution Molal solution: one mole of solute in 1000 g of solution 3- Mole fraction (x): the ratio of the numbers of moles (solute or solvent) to the total number of moles of solution . IF : Number of solute moles : n , Number of solvent moles: N Mole fraction of solvent= N/(n+N) Mole fraction of solule= n/(n+N) The sum = N/(n+N) + n/(n+N) = 1 4- Mass percent: % by number of grams of solute in 100 g of solution. e.g., 20 % of KI means dissolving 20 g of KI in 80 g of H2O.

4. Example 1 : when 10 gm of NaOH are dissolved in 90 gm of water , the density of solution at 20 c° is 1.1089 g/cm3 . Illustrate the four methods of expressing the composition . (M.Wt of NaOH = 40) Answer: 1-Mass % =10% NaOH, 10g NaOH are contained in 100 g of solution. • 2-Mole fraction: • 10 g of NaOH is = 10/40= 0.25 moles (n) • 90 g of H2O is =90/18= 5 moles (N)

5. 3-Molality: moles of solute in 1000g of solution 0.25 mole of NaOH are dissolved in 90 g H2O X 1000g 4-Molarity:moles of solute in one liter of solution molar

6. Types of Solutions • Solutions of gases: (gaseous mixture): Gases mix with one another in all proportions. The air we breathe is a good example of a gaseous mixture. The total pressure of a mixture of gases is the sum of the partial pressures of its components [Dalton’s Law] • e.g. • Solutions of gases in liquids: • Unlike the solubility of one gas in another, the solubility of a gas in a liquid is limited. • The solubility of gases in liquids depends on four factors: • Nature of the gas. • Nature of the solvent • Temperature: the solubility of gases in liquids is accompanied by evolution of heat. i.e., lowering T increase solubility according to Le- Chatelier. • Pressure: affects slightly soluble gases by Henry’s Law.

7. Henry’s Law: it is described the relationship between gas pressure and the concentration of dissolved gas . “At constant temperature, the mass of the gas dissolved by a given volume of liquid solvent is directly proportional to the pressure of the gas above the liquid”. If : m: the mass of gas dissolved by unit volume of the liquid. P: the pressure of the gas above the liquid. • Solutions of gases in solids: There are several ways by which a gas may be taken by a solid: • The gas may be dissolve in the solid & form a homogeneous mixture • The gas may form a chemical compound with the solid e.g. CO2 & CaO. • The gas adheres to the surface of the solid (adsorption of the gas by the solid).

8. The adsorption of a gas by a solid depends on four factors: • Nature of gas: the most easily liquefied gases are readily adsorbed • (e.g. NH3, CO2, and SO2). • Nature of solid: much gas is adsorbed when the solid is in a porous or very finely • divided state and thus possesses a large surface area for example Coal. • Temperature: Adsorption is generally accompanied by evolution of heat so T↓. • Pressure: the mass of gas adsorbed by a given mass of solid increases • (directly proportional) as the pressure is increased. P ↑ adsorption ↑ • If : x: the mass of adsorbed gas, m: the mass of solid. • P: the pressure. • K, n: constants for each temperature, adsorbate, adsorbent

9. Solutions of solids in Liquids: • The solubility of solid in liquids depends upon the nature of both solid and liquid. • When a liquid has dissolved the entire solid, at certain temperature, it will be in • equilibrium ,with an excess of the solid, the solution is said to be saturated. • The solubility of the saturated solution in a solvent under a given conditions is measured by the concentration of the saturated solution. It depends on : • Nature of solid. • Nature of liquid. • Temperature : as T ↑ S↑

10. Binary Liquid Mixtures: • Completely miscible liquids. “totally soluble in each other. When they mix they form one layer” • Completely immiscible liquids. • Partially miscible liquids. Rauolt’s Law: • Relation between partial pressure of pure component and vapor pressure of their mixtures. • “Each liquid will have a partial pressure proportional to its mole fraction in the mixture • If: PA & PB: the vapor pressures of pure liquids A & B. NA& NB: their mole fractions NA = no. of moles of A / total no. of moles P (total vapor pressure) = PA’ + PB’

11. Ideal solutions If the A-A& B-B attraction remains the same after mixing (A, B do not attract each other). • The lie partial pressure of both liquids = line segments between vapor pressure of pure liquid & 100% composition of other liquid. 2. Pressure at any composition lies at line between two pure partial pressure. *PA=Vapor pressure of pure A *PB =Vapor pressure of pure B *Partial vapor pressure of liquid A=Xm. *Partial vapor pressure of liquid B=Xn. *Total pressure =Xo “Geometrically Xn = mo “ Note : the composition is mole fraction Vapor pressure of solution For completely miscible liquid solution

13. ASSIGEMENT: Solve this example graphically