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26-2-2011

26-2-2011. Example. A MgSO 4 (FW= 120.4 g/mol) aqueous solution has a weight fraction of 0.2. What is the molality of the solution? Solution A 0.2 weight fraction means that the solution contains 20 g MgSO 4 and 80 g water (0.080 kg). Mol MgSO 4 = 20 g/(120.4 g/mol) = 0.166 mol

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26-2-2011

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  1. 26-2-2011

  2. Example A MgSO4 (FW= 120.4 g/mol) aqueous solution has a weight fraction of 0.2. What is the molality of the solution? Solution A 0.2 weight fraction means that the solution contains 20 g MgSO4 and 80 g water (0.080 kg). Mol MgSO4 = 20 g/(120.4 g/mol) = 0.166 mol Molality = 0.166 mol/0.080 kg = 2.08 m

  3. Example A CaCl2 (FW = 111.0 g/mol) solution is 4.57 m. Find the mole fraction of CaCl2 and water in solution. Solution 4.57 m means 4.57 mole CaCl2 in 1000 g of water Number of moles of water = 1000 g/(18.0 g/mol) = 55.6 mol XCaCl2 = 4.57 mol/(4.57 + 55.6) = 0.076 Xwater  = 55.6/(4.57 + 55.6) = 0.924

  4. Conversion between Units m = mole/kg 1000 g = 1 kg mass solvent + add molar mass mass solute moles solute mass solution M = mole/L density volume solution 1000 mL = 1 L

  5. Conversion between conc. units Calculate the molarity of a 0.396 m glucose (FW = 180 g/mol) if the density of the solution is 1.16 g/mL. Solution • First calculate the mass of the solution • Use density to change the mass into volume of solution • Calculate M

  6. 0.396 m means 0.396 moles in 1000 g water • Mass of solution = mass of glucose + mass of water • Mass of glucose = mol * FW = 0.396 *180 = 71g • Mass of solution = 71 + 1000 = 1071g • Molarity = (0.396 mol glucose/1071 g soln)* (1.16 g soln/mL soln)* (1000 mL soln/1 L soln) = 0.429 M

  7. From M to m The density of a 2.45 M aqueous methanol (FW = 32g/mol) soln is 0.976 g/mL. What is the molality of soln? Solution • Using density get the mass of the 1 L solution • Get the mass of the solvent by subtracting the mass of methanol from the mass of the soln. • Find the molality

  8. g methanol = M * FW = 2.45*32 = 78.4g Mass of 1 L soln = 1000 ml soln * (0.976g soln/mL soln) = 976 g Mass of water = 976 – 78.4 = 898 g m = 2.45 mol methanol/0.898 kg water) = 2.73 m

  9. moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution 5.86 moles C2H5OH = 0.657 kg solvent What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? • Assume 1 L of solution: • 5.86 moles ethanol = 270 g ethanol • 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg = 8.92 m 12.3

  10. Convert % mass to Molarity What is the Molarity of a 95% acetic acid solution? (density = 1.049 g/mL) If you assume 1 L, that amount of solution = 1049 g 95% of the solution is acetic acid 1049 g solution x 0.95 = 997 g solute 997 g X 1 mol/60.05 g = 16.6 mol solute Since we assumed 1 L, that’s 16.6 mol / 1 L or 16.6 M

  11. Effect of pressure on solubility Liquids and solids exhibit practically no change of solubility with changes in pressure. Gases as might be expected, increase in solubility with an increase in pressure. Henry's Law states that: The solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution. If the pressure is increased, the gas molecules are "forced" into the solution since this will best relieve the pressure that has been applied.The number of gas molecules above solution is decreased.

  12. The solubility of a gas (Cg in mol/L) in a liquid is directly proportional to the pressure of the gas (in atm) above the solution. Cg = kg Pg kg  is the Henry's law constant. This relation is also useful in calculation of the solubility of a gas at one pressure provided that the solubility at some other pressure is known: Cg1 / Cg2 = Pg1 / Pg2

  13. Example The solubility of nitrogen gas at 25 oC and 1 atm is 6.8*10-4 M. What is the concentration of nitrogen dissolved in water under atmospheric conditions? The partial pressure of nitrogen in the atmosphere is 0.78 atm. Solution First find Henry’s constant:

  14. Cg = kg Pg 6.8*10-4 = kg * 1 6.8*10-4 = kg The solubility of nitrogen gas can then be calculated: Cg = kg Pg Cg = 6.8*10-4 * 0.78 Cg = 5.3*10-4 M

  15. Example At 25 oC, O2 gas collected over water at a total pressure of 1.0 atm is soluble to the extent of 0.0393 g/L. What would its solubility be if its partial pressure over water was 800 torr. Pwater = 23.8 torr at 25 oC

  16. PT = PO2 + Pwater 760 torr = PO2 + 23.8 PO2 = 736 torr Substitution in the equation Cg1 / Cg2 = Pg1 / Pg2 gives: 0.0393/CO2 = 736/800 CO2 = 0.0427 g/L For Henry’s law to be valid, the gas should not react with solvent.

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