CHM1046 EXAM 2 12 3 12 4 12 5 Molality m solute moles solvent kg Molarity M solute moles solution L Mass percent mass of component total mass of solution 100 Mole fraction X moles of component total moles making up solution Effect of temperature on solubility of solids and gases o solid can do both As temp o gas As pressure o gas the solution c kP o no change with solids and liquids Henry s Law solubility of gas in a liquid is proportional to the pressure of the gas over c concentration M of dissolved gas P pressure of the gas over the solution k constant mol L atm that depends only on temperature EX the solubility of Nitrogen gas N2 at 25 C and 1 atm is 6 8 x10 4mol L Calculate the molarity when partial pressure of N2 is 0 78 atm use P 1 atm c 6 8 x10 4mol L atm k c P 6 8 x10 4mol L 1atm 6 8 x10 4mol L atm c kP 6 8 x10 4mol L atm 0 78 atm 5 3 x10 4 M 12 6 Colligate Properties of Nonelectrolytes properties that depend only on the amount of solute number of solute particles not the nature of the solute particles used to determine molar mass of solutes BP n BP FP n FP n Vapor Pressure Lowering VP is lowered by addition of solute 2 VP lowering 1 for solute 1 for solvent for solutions that contain 2 or more components obey Raoult s Law o 2 types volatile has VP o PA PA 0 XA PB PB 0 XB Ptot PA PB o Two conditions ideal nonideal o liquids that obey Raoult s Law Ideal solutions o Ideal calculations intermolecular forces b w A B are the same as A A and B B o ex dispersion forces dispersions forces dispersion forces H2O H bond Ethanol H bond H bonds o Nonideal no calculations intermolecular forces b w A B are different from A A and B B Deviation from Raoult s Law PT is predicted by Raoult s Law AB AA PT is predicted by Raoult s Law AB AA BB BB nonvolatile does not has VP solid Nonvolatile relationship b w VP solute moles is described by Raoult s Law P1 X1 P1 0 o P1 VP of solution o P1 0 VP of PURE solvent o X1 mole fraction of the solvent if solution contains only one solute o X1 1 X2 o P1 0 P1 P X2 P1 0 o X2 mole fraction of solute EX Calculate the VP of a solution made by dissolving 218g glucose molar mass 180 2 g mol in 460mL of water at 30 C What is the VP lowering The VP of pure water at 30 C is given in Table 5 3 p199 31 82 mmHg Assume the density of the solvent is 1 00 g mL a 218g solute 460mL solvent H2O P1 0 31 82 mmHg b X1 H2O moles mol of solute mol of solvent c 218g 1mol 180 2g 1 21 moles moles H2O V D M 460mL 1g mL 460g d 460g 1mol 18 02g 25 5 moles X1 25 5 mol 25 5 mol 1 21 mol 0 955 e P1 X1 P1 0 0 955 31 82 mmHg 30 4 mmHg VP f P P1 0 P1 31 82 mmHg 30 4 mmHg 1 4 mmHg VP Lowering Boiling Point Elevation o normal BP is temp at which VP is to atm press o adding nonelectrolyte solution BP o adding a solute to solvent BP solute should be nonvolatile Tb Kb m Tb Tb Tb 0 Tb 0 BP of pure solvent Tb BP of solution C m molality molal of solution mol kg Kb molal BP elevation constant C m ALWAYS Tb 0 Nonelectrolytes Tb Kb m Electrolytes Tb Kb m i EX Calculate the boiling point of a solution made by dissolving 50 0 g of sucrose in 200 0 g of water This time we re given a boiling point elevation problem Here s the formula for the boiling point of a solution Tb Tb Tb water boils at 100 C 1 Tb BP of solution Tb 0 BP of solvent Tb BP elevation 2 Tb Kb m Tb Kb mass molar mass mass of solvent kg Tb 0 512 C m 50 0g 342 30g mol 0 2 kg Tb 0 374 C 3 Tb Tb 0 Tb Tb 100 C 0 374 C Tb 100 374 C BP Freezing Point Depression o any solute volatile or nonvolatile will decrease the FP of the solvent water 0 C add sugar FP 0 C o Tf Kf m Tf FP depression Kf molal FP depression constant C m m molality molal of solution mol kg Nonelectrolytes Tf Kf m Electrolytes Tf Kf m i EX Ethylene glycol C H O is a common automobile antifreeze It is water soluble and fairly nonvolatile Boiling point 197 C Calculate the freezing point of the solution containing 651 g of this substance in 2505 g of water water freezes at 0 C Here is the equation Tf Tf 0 Tf 1 Tf FP solution Tf 0 FP solvent Tf FP depression 2 Tf Kf m Kf molal FP depression constant m molality of solution 3 Tf Kf m Tf Kf mole of solute mass of solvent kg Tf Kf mass molar mass mass of solvent kg Tf 1 86 C m 651g 62 07g mol 2 505kg Tf 7 79 C 4 Tf Tf 0 Tf Tf 0 C 7 79 C Tf 7 79 C FP Osmotic Pressure o amount of pressure necessary to cause osmosis to stop amount of pressure necessary to achieve an equilibrium passage through semi permeable membrane o electrolyte i M R T o nonelectrolyte M R T EX What is the osmotic pressure of a solution prepared by adding 13 65 g of sucrose C12H22O11 to enough water to make 250 mL of solution at 25 C 1 i M R T osmotic press in atm I van t Hoff factor of solute M molar con mol L R 0 0821 L atm mol K T absolute temp K 2 13 65g 1 mol 342g 0 04 mol 3 0 04 mol 250 mL 1 L 1000 mL 0 04 mol 0 25 L 0 16 mol L 4 T 25 C 273 298 K 5 Sucrose does not dissociate in water so the van t Hoff factor 1 6 i M R T 1 0 16 mol L 0 0821 L atm mol K 298 K 3 9 atm osmotic press Three types of solution 12 7 o Isotonic conc outside conc inside membrane ex blood cells o Hypertonic more conc outside of cell causing fluid to leave the cell cell shrinks o Hypotonic more conc inside cell fluid rushes in causing cell to swell burst osmotic lysis Electrolytes substances that produce ions in solutions ex NaCl Nonelectrolytes does not dissociate at all in solution does not produce any ions ex glucose Colligative Properties depend only on the …
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