An aqueous solution of ethanol, C 2 H 5 OH, is 0.05 mol fraction in solute.

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1 CHEM 101/105 DISC 26-NOV-02 When 3.24 grams of mercuric nitrate, Hg(NO 3 ) 2, are dissolved in 1000 grams of water, the freezing point of the solution is found to be deg.c. When grams of mercuric chloride, HgCl 2, are dissolved in 1000 grams of water, the freezing point of the solution is deg.c. Are either of these salts dissociated into ions in aqueous solutions? An aqueous solution of ethanol, C 2 H 5 OH, is 0.05 mol fraction in solute. Express the concentration of this solution in units of molality ( m ), molarity ( M ), and percent ( % ). The density of the solution is g/ml. A 7.00 % aqueous ammonium sulfate solution (solute = (NH 4 ) 2 SO 4 ) has a density of g / ml. Express the concentration of this solution in unit of molality ( m ), molarity ( M ), and mol fraction ( χ ). Determine the freezing point temperature of this solution.

2 1. Are either of these salts dissociated into ions in aqueous solution? a. the first solution with Hg(NO 3 ) 2 as solute: calculate EXPERIMENTAL molality from the freezing point. T m = = deg. C K 186. deg. C / m = 0.03 calculate molality based on formula of Hg(NO 3 ) 2 : Formula Weight Hg(NO 3 ) 2 = (14+48) = 324 g / mole m = = 324. g / ( 325 g / mole) 10. kg water = 0.01 Compare molality based on F.P. Temperature LOWERING, to that based on formula of Hg(NO 3 ) 2 The F.P. LOWERING molality is THREE TIMES that expected from the formula. Conclusion: there must be THREE times the number of items present in solution to cause this three-fold increase. What is it about the solute Hg(NO 3 ) 2 that might result in a THREE-fold increase in number of items in solution? Conclusion: if it were completely dissociated in ions, as shown: Hg(NO 3 ) 2 Hg 2+ (aq) + 2 NO 3 1- (aq) ONE formula unit YIELDS THREE IONS then the three-fold change in freezing point temperature actually experienced could be accounted for. FINAL CONCLUSION: solution. Hg(NO 3 ) 2 is completely ionized in aqueous b. the second solution with HgCl 2 as solute: calculate EXPERIMENTAL molality from the freezing point. T m = = deg. C K 186. deg. C / m = 0.04

3 calculate molality based on formula of HgCl 2 : Formula Weight HgCl 2 = (35.45) = g / mole m = = g/(271.5 g/mole) 1.0 kg water = 0.04 Molality from experimental F.P. Temperature lowering, and that based on formula HgCl 2 unit are the same. Conclusion: HgCl 2 IS NOT IONZIED in aqueous solution. COLLIGATIVE PROPERTIES depend on the NUMBER of solute items present in solution. Colligative properties involve i. lowering of vapor pressures, ii. increase in B.P. temperature, iii. decrease in F.P. tempera rue, iv. osmotic pressures, *****Colligative effects of IONIC SOLUTES will be greater than predicted by chemical formula if the solute is dissociated into ions. (return to Lect-17 handout...) Walls of living cells are sort of like SPM. Cell walls can burst when immersed in solutions of lower concentrations, or shrink when immersed in solutions of higher concentrations. In RBC, cell walls are at an osmotic pressure of about 7.7 atm, measured against solvent water. Aqueous solutions to be introduced into the blood stream must have the same osmotic pressure as RBC. Otherwise Such solutions are said to be isotonic with blood. (j) What are the molarities, at 25 deg C, of an isotonic lactose solution? of an isotonic saline solution? A solution isotonic with blood will also have an osmotic pressure of 7.7 atm. What is the molarity concentration of such a solution? Π = MRT so M = 7. 7atm L atm ( K) K mole M(solution isotonic with blood) = molar a. Lactose ( C 12 H 22 O 11 ) is a molecular sugar. It does not ionize in aqueous solution. So the a molar solution of lactose would have an osmotic pressure of 7.7 atm. and be isotonic with blood. Solutions of glucose (another molecular sugar) are often used in hospital settings. A Molar solution of glucose would be isotonic with blood.

4 b. Sodium chloride, NaCl, is an ionic substance *****. It is COMPLETELY IONZIED in aqueous solution with each NaCl formula unit dissociated into TWO ions. So colligative properties of a sodium chloride solution would have TWICE the effects expected of a molar solution. Te osmotic pressure of a M sodium chloride solution is 15.4 atm. A slide was shown illustrating the effect on red blood cells when such a non-isotonic solution ( a hypertonic solution in this case) is administered. The cell wall (acting as a semi-permeable membrane) permits intracellular water to escape from the cell in an attempt to dilute the administered solution. RBC's collapse and die (crenation). If a solution were administered having a lower osmotic pressure than RBC's (a hypotonic solution) then extracellular water would enter the cell causing swelling that could lead to cell rupture. Either case is to be avoided at all times. Consequently, isotonic saline solutions have molarities of (0.303 / 2) or Molar. Convert this molarity to percentages (the density of an isotonic NaCl solution is g/ml): ONE LITER of this solution has a mass of 1,006.5 grams, and contains moles of NaCl. mass of moles NaCl = ( moles) x ( 58.5 g / mole ) = 8.86 grams NaCl 886. Per Cent = part / whole = grams NaCl g solution x 0.88 % = Concentrations of isotonic saline solutions used in hospital settings are labeled as 0.9 %. Express the concentration of a glucose ( C 6 H 12 O 6 ) solution isotonic with blood as percent (the density of this solution is g/ml): ONE LITER of this solution has a mass of 1,020.6 grams, and contains moles of glucose. mass of moles of glucose = ( moles) x ( 180 g / mole) = g glucose. Per Cent = part / whole = g glucose x100 = 5.3 % g solution (return to second problem on today's handout sheet Given mol fraction ethanol (solute) = 0.050: If only two components in this solution ( solute and solvent ), then mol fraction solvent = = 0.95 Find mass of moles ethanol = ( moles) x ( 46 g / mole ) = 2.3 grams ethanol (solute) Find mass of 0.95 moles water = ( 0.95 moles ) x ( 18 g / mole ) = 17.1 grams water (solvent)

5 (i) Per Cent ( by weight) part / whole = 23. g solute = 11.9 % ( ) g solution (ii) molality = = = 2.92 molal (iii) Given density of solution, then ONE LITER of solution must have a mass of 997 grams. Of this 997 grams solution, 11.9 % is solute. ( 11.9 % of 997 = 119 grams solute) How many are present in ONE LITER solution? = 119 grams / 46 g / mole = 2.58 moles / Liter solution = 2.58 M 3. Given Per Cent composition and density... ONE LITER of this solution has a mass of 1,044.5 grams, of which 7 % is solute. Mass solute = ( part solute ) x ( 1,044.5 grams whole solution ) = 73.1 grams solute Moles solute = 73.1 g / 132 g / mole = Mass solvent = 1, = grams solvent Moles solvent = g / 18 g / mole = (i) molality = = = 0.57 molal (ii) mol fraction = total moles = ( ) = solute (iii) Molarity = { ONE LITER of solution contains 73.1 g or moles of solute) M = Molar (iv) What is the freezing point temperature of this solution? T = K f m = ( 1.86 deg / molal ) ( 0.57 molal) X (3 ions / formula unit)***** T = 3.18 deg.c (lowering of F.P. temp) Actual Freezing Point Temperature of solution = 3.18 deg.c