5 Key Terms #2a solute: the substance that will be dissolved. Ex: NaCl(s) #2b solvent: substance that will do the dissolving. Ex: H 2 O(l)
6 #4 Polar dissolve other polars Ex: H 2 O and NaCl Nonpolar dissolves other nonpolars Motor oil and gasoline
7 #15a Dissociation- separation of ions that occurs when an ionic compound dissolves NaCl(s) H2O Na + (aq)+ Cl - (aq)
8 #10 Electrolyte- any substance that dissociates into ions when dissolved in a suitable medium or melted. forms a conductor of electricity. NaCl is a strong electrolyte, meaning that most of its bonds will break to form ions. As opposed to a weak electrolyte and its few bonds that break into ions.
9 #12 molality (m): number of mole of solute amount of solvent (kg) NOT the same as #2d or #11 molarity (M) M= m= the mass of the solute will not change with varying temperatures, while the volume of the solvent will. Therefore, for our experiment, molality would provide a constant mass to work with. Number of moles of solute Amount of solvent (L) Number of mole of solute Amount of solvent (kg)
10 #16 Colligative properties- any property of a solution that is changed by the addition of a solute. Ex: adding NaCl to water increases the boiling point.
11 #17b dealing with colligative properties: Electrolytes in solutions. As stated before, NaCl, an electrolyte, increases boiling point when it is added to water. How to see this mathematically requires the use of a formula.
12 Δt = mK b ( ) Δ delta (change) t temperature m molality K b constant. For water, it is.51 C/m Moles of ions Per electrolyte in solution
13 Δt = Δt: we are solving for the change in temp. to see if increasing amounts of NaCl will increase the boiling point. Our first amount of NaCl that we tested was 15g
14 m molality: number of mole of solute amount of solvent (kg) Step 1: find number of mole of solute. Use dimensional analysis 15 g NaCl 1 mol NaCl 58.443 g NaCl.257 mol NaCl
15 m cont. Step 2: find amount of solvent (kg) * remember that 1 mL of water = 1 g of water 50 mL H 2 O = 50 g H 2 O =.050 kg H 2 O Step 3: plug back in to find m. 257 mol NaCl.050 kg H 2 O 5.140 mol/kg m
17 How many moles of ions are there for every electrolyte? (think back to dissociation) NaCl(s) From 1 mol NaCl(s), the reaction yields 1 mol Na (aq) and 1 mol Cl (aq). So for every electrolyte in the solution, you have 2 mols of ions. Moles of ions Per electrolyte in solution H2OH2O Na + (aq)+ Cl - (aq)
18 Plugging the numbers in. Δt = mK b ( ) Δt= (5.140)(.51)(2) Δt= 5.243 we should find that the boiling point of the solution increases by about 5°C when 15 g NaCl is added Moles of ions Per electrolyte in solution
19 Your Turn! Attempt to find the change in temperature for an added 30 g NaCl.
20 Your turn! Remember the formulas: Δt = mK b ( ) m= K b =.51 C/m Moles of ions Per electrolyte in solution Number of moles of solute Amount of solvent (kg) moles of ions Per electrolyte in solution 2
21 30 g NaCl 1 mol NaCl 58.443 g NaCl K b =.51 C/m 2 moles of ions Per electrolyte in solution 10.260 mol/kg.513 mol NaCl.050 kg H 2 O
22 Δt = mK b () Δt = (10.266) (.51) (2) Δt = 10.471 The new boiling point temperature should increase by about 10ºC. Moles of ions Per electrolyte in solution
23 With 43g NaCl Δt = mK b ( ) Moles of ions Per electrolyte in solution Δt = (14.715)(.51)(2) Δt =15.009
24 Conclusion An increased amount of NaCl solute in the tap water will result in an increase in the boiling point of the solution.