Chap13_Part4

# Chap13_Part4 - 1) Mass Percentage (wt) ...

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Unformatted text preview: 1) Mass Percentage (wt%)  For dilute solu,ons, some,mes the mass % is referred to   as “parts per million”:  mass of component ppm = x 10 6 total mass of solution For very dilute solu,ons, some,mes the mass % is referred to   as “parts per billion”:  mass of component ppb = x 10 9 total mass of solution For dilute and very dilute solu,ons, the total mass  of the solu,ons is ≈ the total mass of the solvent.  2  2) Mol Frac5on  mols of component mol fraction of component = X = total mols in solution 3) Molarity  mols of solute molarity = M = liters of solution 4) Molality  mols of solute molality = m = kg solvent When the solvent is water, 1 L ≈ 1 kg.  Thus, when the solu,on is  dilute, the denominators of 3) and 4) are equivalent.  3  Calcula5ng Molarity and Molality   If we know the density  of the solu,on, we  can calculate the  molality from the  molarity and vice  versa.  4  A solu,on containing 80.5 gm of vitamin C (C6H8O6) in 210 gm water has a  density of 1.22 gm/mL. Calculate the (a) mass percentage; (b) mol frac,on;   (c) molality; and (d) molarity.    (a)  5  A solu,on containing 80.5 gm of vitamin C (C6H8O6) in 210 gm water has a  density of 1.22 g/mL. Calculate the (a) mass percentage; (b) mol frac,on;   (c) molality; and (d) molarity.    (a)  mass solute 80.5 x100 = = 27.7% mass solution 80.5 + 210 (b)  6  A solu,on containing 80.5 gm of vitamin C (C6H8O6) in 210 gm water has a  density of 1.22 gm/mL. Calculate the (a) mass percentage; (b) mol frac,on;   (c) molality; and (d) molarity.    (a)  (b)  mass solute 80.5 x100 = = 27.7% mass solution 80.5 + 210 molecular weight solute = 6(12) + 8(1) + 6(16) = 176 mol solute = gm mol 80.5 gm = 0.457 mol 176 gm / mol molecular weight solvent = 18 gm mol 210 gm = 11.7 mol 18 gm / mol 0.457 mol fraction = = 0.038 11.7 + 0.457 mol solvent = 7  A solu,on containing 80.5 gm of vitamin C (C6H8O6) in 210 gm water has a  density of 1.22 gm/mL. Calculate the (a) mass percentage; (b) mol frac,on;   (c) molality; and (d) molarity.    (c)  8  A solu,on containing 80.5 gm of vitamin C (C6H8O6) in 210 gm water has a  density of 1.22 gm/mL. Calculate the (a) mass percentage; (b) mol frac,on;   (c) molality; and (d) molarity.    (c)  mol solute = 0.457 mol mass solvent = 210 gm = 0.210 kg 0.457 mol mol molality = = 2.18 0.210 kg kg (d)  9  A solu,on containing 80.5 gm of vitamin C (C6H8O6) in 210 gm water has a  density of 1.22 gm/mL. Calculate the (a) mass percentage; (b) mol frac,on;   (c) molality; and (d) molarity.    (c)  mol solute = 0.457 mol mass solvent = 210 gm = 0.210 kg 0.457 mol mol molality = = 2.18 0.210 kg kg (d)  mol solute = 0.457 mol (210 + 80.5) gm volume solution = = 238 mL = 0.238 L 1.22 gm / mL 0.457 mol mol molarity = = 1.92 0.238 L L 10  Colliga5ve Proper5es  Changes in colliga,ve proper,es depend only on the number of   solute par,cles present, not on the iden)ty of the solute par,cles.  These proper,es are called “colliga,ve”:     1) vapor pressure lowering     2) boiling point eleva,on     3) freezing point depression     4) osmo,c pressure  Lowering the Vapor Pressure  A liquid in a closed container will come into  equilibrium with its vapor.  At that point, the gas pressure is called the  Vapor pressure.  Substances that have no measurable vapor   pressures are nonvola,le; substances that   do are vola,le.  The vapor pressure of a vola,le solvent containing  a nonvola,le solute is lower than the vapor   pressure of the pure solvent.  The amount by which the vapor pressure is   reduced depends on the concentra,on of solute.  The solute pulls solvent into the solu,on.  Raoult’s Law:  Vapor Pressure Decrease with Solute  “The vapor pressure of a gas above a solu,on equals the product of the mol   frac,on of solvent in the solu,on and the vapor pressure of the pure solvent”.  PA = X A PA 0 PA = vapor pressure of   gas above solu,on   XA = mol frac,on of   solvent in solu,on  PA0 = vapor pressure  of pure solvent  XA usually does not diﬀer too much from one and the eﬀects are usually small.  Francois‐Marie Raoult (1830‐1901)  ‐  French Chemist  ‐  Raoults law  ‐  Work on freezing point depression  ‐ Lowering the vapor pressure  ‐ PhD work on electrochemistry  The vapor pressure of pure water at 70oC is 0.31 atm.  Calculate the vapor   pressure of 125 gm water containing 35.0 gm glycerin (C3H8O3) in water   at this temperature.   The vapor pressure of pure water at 70oC is 0.31 atm.  Calculate the vapor   pressure of 125 gm water containing 35.0 gm glycerin (C3H8O3) in water   at this temperature.   molecular weight of glycerin = 3(12) + 8(1) + 3(16) = 92 35.0 gm mol solute = = 0.38 mol 92 gm / mol 125 gm mol solvent = = 6.94 mol 18 gm / mol 6.94 mol fraction solvent = = 0.95 6.94 + 0.38 0 PA = X A PA = (0.95)(0.31 atm) = 0.29 atm gm mol Although the eﬀects are small, the lowering of vapor pressure can be used to determine solute  concentra,ons.  At the top, the solvent is 1 kg ≈ 1 L H2O and the height diﬀerence is 17.54 mm.    At the bobom, a certain amount of the sugar mannitol (182.2 gm/mol) has been added and   the height diﬀerence is 17.45 mm (requiring something other than a ruler to measure).    Assume that the volume does not change.  Find the concentra,on of mannitol in units of gm/L.  barometer = closed top  Although the eﬀects are small, the lowering of vapor pressure can be used to determine solute  concentra,ons.  At the top, the solvent is 1 kg ≈ 1 L H2O and the height diﬀerence is 17.54 mm.    At the bobom, a certain amount of the sugar mannitol (182.2 gm/mol) has been added and   the height diﬀerence is 17.45 mm (requiring something other than a ruler to measure).    Assume that the volume does not change.  Find the concentra,on of mannitol in units of gm/L.  XA = PA PA 0 = 17.45 mm = 0.9949 17.54 mm 1000 gm = 55.56 mol 18 gm / mol n solvent XA = ⇒ X A (n solvent + n solute ) = n solvent n solvent + n solute n solvent = ⇒ barometer = closed top  ⇒ n solute = (1 − 0.9949)(55.56 mol ) = 0.2848 mol 0.9949 182.2 gm = (0.2848 mol )( ) = 51.89 gm mol gm ⇒ c solute = 52 L n solute = m solute X A n solute = (1 − X A )n solvent (1 − X A )n solvent XA ‐  Solu,ons that obey Raoult’s law : Ideal Solu,ons  ‐  Devia,ons from the Ideal behavior of the solu,ons: Based on    intermolecular forces.   ‐ Strong Solvent/Solvent  or Solute/Solute interac,ons: Vapor   pressure is more than predicted by Raoult’s Law.  (Example: Actual P= 1 atm, and calculated P according to Raoult’s law ager the   Addi,on of solute is 0.95 atm‐ you will ﬁnd PA>0.95 but < 1)  ‐ Strong Solute/Solvent interac,ons: Vapor pressure is less than   predicted by Raoult’s law.  (Example: Actual P= 1 atm, and calculated P according to Raoult’s law ager the   Addi,on of solute is 0.95 atm‐ you will ﬁnd ﬁnd PA<0.95)  IMPORTANT: EXAM  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## This note was uploaded on 09/13/2011 for the course CHEM 102 taught by Professor Todd during the Spring '08 term at UNC.

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