Major Points*a solution is a homogeneous mixture
*parts of a solution: solute, solvent *solubility *saturated and unsaturated *concentrated solution and dilated solution *mass percent *Polar compound: Polar compound will dissolve in water - compounds that will contain charges ( + or -) - Not all things are soluble in water. ActivityVideos |
Online ResourcesMolarityhttp://www.occc.edu/kmbailey/Chem1115Tutorials/Molarity.htm
Molarity and Stoichiometry: http://leifchemistry.blogspot.com/2011/04/molarity-and-stoichiometry.html Example problems1. We have 28 psi convert to atm:
28 psi x 1 atm/14.7 psi= 1.9 atm 2. We have a 1.5 L sample of Freon-12 with a pressure of 56 torr. The pressure changed to 150 torr, what is the new volume if temperature is constant? V1=1.5 L; P1=56 torr; P2=150 torr; V2=? PV=PV (56 torr) (1.5L)/150 torr=0.56 L 3. V/T=V/T We have a 2.0 L sample of air that is collected at 298K and the cooled to 278 K. Pressure is constant. What is the new volume? Vi=2.0 L; Ti=298 K; Tf=278K; Vf=? 2.0L/298K x 278K=1.9L 4. P1V1/T1=P2V2/T2 A sample of gas has a volume of 283 mL at 298 K and 0.500 atm pressure. What is the volume at 373 K and 1.00 atm? T1=298 K; V1=283 mL; P1=0.500 atm; T2=373 K; V2=?; P2=1.00 atm (0.500 atm x 283mL x 373K)/(298 K x 1.00 atm)=177mL 5. V1/n1=V2/n2 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? V1=5.00 L; n1=1.80 mol; V2=?; n2=0.965 mol (5.00 L x 0.965mol)/1.80 mol=2.68L 6. PV=nRTAt what temperature will 0.654 mole of neon gas occupy 12.30 L at 1.95 atm? n=0.654mole; V=12.30L; P=1.95 atm; R=0.08206 (1.95x 12.30)/(0.654 x 0.08206)=447K 7. One gas has a pressure of 40 atm, another of 60 atm, and a third of 75 atm. If combined what would the total pressure be? P1=40 atm, P2=60 atm, P3=75 atm Total pressure=175 atm 8. A solution is prepared by dissolving 1.0 g of sodium Chloride in to 48 g of water. What is the mass percent of sodium? mass of solute=1.0 g mass of solution=1.0 + 48=49 1.0/49 x 100=2.0% 9. We have 0.521 moles of Ne and a volume of 0.125 L. Find the Molarity. 0.521mol/0.125 L=4.17M 10. How much 4.5 M KCl solution would you need to make 750.mL of 0.15M KCl solution? CfVf=CiVi Cf=0.15M; Ci=4.5 M; Vi=?; Vf=750.mL (0.15M x 750.mL)/4.5M=25mL 11. What volume of 0.20M HNO3 is needed to react completely with 37g of Ca(OH)2? 2HNO3 + Ca(OH)2 -> 2H2O + Ca(NO3)2 37 g Ca(OH)2 x 1 mol Ca(OH)2/74.0918 g Ca(OH)2 x 2 mol HNO3/1 mol Ca(OH)2 x 1 L HNO3/0.20 mol HNO3=5.0L HNO3 |