1. Ch. 11 Solutions
11.1 Solution Composition

2. 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

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

4. Molarity
number of moles of solute per L or solution

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

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

7. Molality
number of moles of solute per kg of solvent

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

9. 11.2 Energies of Solution Formation
Ch. 11 Solutions
11.2 Energies of Solution Formation

10. Solubility “like dissolves like” polar dissolves polar
nonpolar dissolves nonpolar

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

12. 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

13. Energy of Solubility Process

14. 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

15. 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 +

16. 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

17. Energy of Solubility Process