Ch. 11 Solutions 11.1 Solution Composition

Composition solute solvent substance being dissolved solvent what is dissolving the solute when both are liquids, the one with the largest volume is the solvent

Example 1.00 g C2H5OH is added to 100.0 g of water to make 101 mL of solution. Find the molarity, mass % mole fraction and molality of ethanol.

Molarity number of moles of solute per L or solution

Mass Percent also called weight percent percent by mass of the solute in the solution

Mole Fraction ratio of number of moles of a part of solution to total number of moles of solution

Molality number of moles of solute per kg of solvent

not covering normality. skip it if it shows up in a HW question

11.2 Energies of Solution Formation Ch. 11 Solutions 11.2 Energies of Solution Formation

Solubility “like dissolves like” polar dissolves polar nonpolar dissolves nonpolar

Solubility Process expand solute molecules expand solvent molecules solute and solvent interact

Energy of Solubility Process Steps 1 and 2 require energy to overcome IMFs endothermic Step 3 usually releases energy exothermic enthalpy of solution sum of ∆H values can be – or + ∆Hsoln = ∆H1 + ∆H2 + ∆H3

Energy of Solubility Process

Case 1: oil and water oil is nonpolar (LD forces) water is polar (H bonding) ∆H1 will be small for typical size ∆H2 will be large ∆H3 will be small since there won’t be much interaction between the two ∆Hsoln will be large and + b/c energy required by steps 1 and 2 is larger than the amount released by 3

Case 2: NaCl and water NaCl is ionic water is polar (H bonding) ∆H1 will be large ∆H2 will be large ∆H3 will be large because of the strong interaction between ions and water ∆Hsoln will be close to zero- small by +

Energy of Solubility Process Enthalpy of hydration - ∆Hhyd combines ∆H2 + ∆H3 NaCl(s)  Na+(g) + Cl-(g) ∆H1=786 kJ/mol H2O(l) + Na+(g) + Cl-(g)  Na+(aq) + Cl-(aq) ∆Hhyd=∆H2 + ∆H3=-783 kJ/mol ∆Hsoln=3 kJ/mol

Energy of Solubility Process