Review Session-5

# Review Session-5 - REVIEW SESSION‐ 5  Proper1es of Solu1ons

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Unformatted text preview: REVIEW SESSION‐ 5  Proper1es of Solu1ons  13.60) (a) What is ideal solu4on? (b) The vapor pressure of pure  water at 60oC is 149 torr. The vapor pressure of water over a solu4on  containing equal number of moles of water and ethylene glycol is 67  torr. Is the solu4on is ideal according to Raoults law? Explain?  13.60) (a) What is ideal solu4on? (b) The vapor pressure of pure  water at 60oC is 149 torr. The vapor pressure of water over a solu4on  containing equal number of moles of water and ethylene glycol is 67  torr. Is the solu4on is ideal according to Raoults law? Explain?  PA = X A PA 0 PA = vapor pressure of   gas above solu4on   XA = mol frac4on of   solvent in solu4on  PA0 = vapor pressure  of pure solvent  13.60) (a) What is ideal solu4on? (b) The vapor pressure of pure  water at 60oC is 149 torr. The vapor pressure of water over a solu4on  containing equal number of moles of water and ethylene glycol is 67  torr. Is the solu4on is ideal according to Raoults law? Explain?  (a)  Ideal solu4on is that obeys Raoults law  PA = X A PA 0 (b)   Number of moles of water = number of moles of Ethylene glycol  XA = 0.5 ,  PA = (0.5) (149) = 74.5mm of Hg  This solu4on has a vapor pressure that is higher than the value  obtained experimentally (67 torr). Theore4cal Pressure (Raoults  law) is more than observed Pressure. This is called a nega4ve  devia4on.  13.62) (a) Calculate the VP of water above a solu4on prepared by  dissolving 32.5g of glycerin(C3H8O3) in 125g of water at 343K. (The VP  of water at 343K = 233.7 torr) (b) Calculate the mass of ethylene  glycol (C2H6O2) that must be added to 1kg of ethanol (C2H5OH) to  reduce the VP by 10 torr? (The VP of ethanol at 35oC is 100 torr)  13.62) (a) Calculate the VP of water above a solu4on prepared by  dissolving 32.5g of glycerin(C3H8O3) in 125g of water at 343K. (The VP  of water at 343K = 233.7 torr) (b) Calculate the mass of ethylene  glycol (C2H6O2) that must be added to 1kg of ethanol (C2H5OH) to  reduce the VP by 10 torr? (The VP of ethanol at 35oC is 100 torr)  (a)  Number of moles of Glycerin = 32.5/92.1 = 0.3529 mol        Number of moles of water = 125/18 = 6.94  mol   Mol frac4on of solvent X water= 6.94/(6.94+0.353) = 0.952  PH2O = 0.952X233.7 = 222 torr    13.62) (a) Calculate the VP of water above a solu4on prepared by  dissolving 32.5g of glycerin(C3H8O3) in 125g of water at 343K. (The VP  of water at 343K = 233.7 torr) (b) Calculate the mass of ethylene  glycol (C2H6O2) that must be added to 1kg of ethanol (C2H5OH) to  reduce the VP by 10 torr? (The VP of ethanol at 35oC is 100 torr)  (a)  Number of moles of Glycerin = 32.5/92.1 = 0.3529 mol        Number of moles of water = 125/18 = 6.94  mol   Mol frac4on of solvent X water= 6.94/(6.94+0.353) = 0.952  PH2O = 0.952X233.7 = 222 torr    (b) X glycerin = 10/100 =0.1 or X ethanol =90/100 =0.9   moles of ethanol 1000g/ 46.07= 21.7 and   “Y” moles of Ethylene glycol    0.1 = Y/Y+21.7  Y= 2.41 mol    Mol.Wt of glycerin = 62.07g/mol  62.07X2.41 = 150g of  ethylene glycol  13.66) Arrange the following aqueous solu4ons, each 10% by mass in  solute, in order of increasing Boiling Point? Glucose (C6H12O6),  Sucrose (C12H22O11), NaNO3.   13.66) Arrange the following aqueous solu4ons, each 10% by mass in  solute, in order of increasing Boiling Point? Glucose (C6H12O6),  Sucrose (C12H22O11), NaNO3.   10%  10g in 100g of solu4on  Number of moles: Wt/ Mwt  Higher the Mwt, Lower the number of moles & Molecules.  Sucrose has highest MWt, least number of Molecules  NaNO3: Electrolyte: can give 2 ions: More molecules  The boiling points of the solu4ons   Sucrose < Glucose < NaNO3    NaNO3 will have high Boiling Point   13.71) How many grams of ethylene glycol (C2H6O2) must be added to  1kg of water to produce a solu4on that freezes at ‐5oC  13.71) How many grams of ethylene glycol (C2H6O2) must be added to  1kg of water to produce a solu4on that freezes at ‐5oC  ΔTf = 5oC = Kf (m)  (m) = 5oC/1.86 oC/m  =2.69 molals of C2H6O2  m = mol of C2H6O2 / kg water  Mol of C2H6O2= m x 1kg water = 2.69X1 =2.69moles of C2H6O2  2.69 X 62.07 = 167g of C2H6O2  13.72) What is the freezing point of an aqueous solu4on that boils at  105oC?  13.72) What is the freezing point of an aqueous solu4on that bolis at  105oC?  Kb =0.51 and Kf =1.86  Bp=105  ΔT=5oC  ΔT = Kb m  m= ΔT/Kb = 5/0.51 = 9.8m  ΔTf = Kf m 1.86 X9.8 = 18oC  Freezing point = 0‐18 = ‐18oC    13.74) Sea water contains 3.4g of salts for every liter of solu4on.  Assuming that the solute consists en4rely NaCl, calculate the osma4c  pressure of sea water at 20oC  13.74) Sea water contains 3.4g of salts for every liter of solu4on.  Assuming that the solute consists en4rely NaCl, calculate the osma4c  pressure of sea water at 20oC  Π =MRT, T = 293K  M= (3.4/58.5X1L) X2 = 0.116M  P = MRT = 0.116 X 0.0821X 293K = 2.8 atm  13.78) A dilute aqueous solu4on of an organic compound is formed  by dissolving 2.35g of the compound in water to form 0.35L solu4on.  The resul4ng solu4on has an osma4c pressure of 0.605atm at  25oC .Assuming the organic compound is non electrolyte, what is the  molar mass?  13.78) A dilute aqueous solu4on of an organic compound is formed  by dissolving 2.35g of the compound in water to form 0.35L solu4on.  The resul4ng solu4on has an osma4c pressure of 0.605atm at  25oC .Assuming the organic compound is non electrolyte, what is the  molar mass?  Π =MRT, T = 298K  M= (0.605/0.0821X 298)= 0.0247M  Molarity = wt/(Molwt X Vol in Liters)  Mol. Wt = Wt/(MolarityX Vol in Liters)  Mol wt = 2.35/(0.0247X0.35) = 272g/mol   13.78) Fish needs at least 4ppm dissolved oxygen for survival. (a) What  is this concentra4on in mol/L (b) What par4al pressure of O2 above the  water is needed to obtain this concentra4on at 10oC? (Henry’s law  constant for O2 at this temperature is 1.71X10‐3mol/L‐atm  13.78) Fish needs at least 4ppm dissolved oxygen for survival. (a) What  is this concentra4on in mol/L (b) What par4al pressure of O2 above the  water is needed to obtain this concentra4on at 10oC? (Henry’s law  constant for O2 at this temperature is 1.71X10‐3mol/L‐atm  (a)  4ppm =  4mg of O2/1kg of solvent                     = 4X10‐3g/ 1L solvent          = 4X10‐3g/ (1L solventX 32g of oxygen)          = 1.25X 10‐4M   13.78) Fish needs at least 4ppm dissolved oxygen for survival. (a) What  is this concentra4on in mol/L (b) What par4al pressure of O2 above the  water is needed to obtain this concentra4on at 10oC? (Henry’s law  constant for O2 at this temperature is 1.71X10‐3mol/L‐atm  (a)  4ppm =  4mg of O2/1kg of solvent                     = 4X10‐3g/ 1L solvent          = 4X10‐3g/ (1L solventX 32g of oxygen)          = 1.25X 10‐4M   (b)  SO2= kPO2   PO2= SO2/k = 1.25X10‐4 mol / 1.71X10‐3 mol/L‐atm   PO2 = 0.0731 atm = 60mm of Hg   REVIEW SESSION : SATURDAY:   CHAPMAN 211 at 10.30 am to Noon   ...
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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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