Chapter 11 Properties of Solutions
Terms:
Molarity- the number of moles of solute per liter of solution
Mass percent- the percent by mass of the solute in the solution
Mole fraction- the ratio of the number of moles of a given component to the total number of
moles of solution
Molality- the number of moles of solute per kilogram of solvent
Normality- the number of equivalences per liter of solution
Enthalpy (heat) of solution- the sum of the H values for the steps
Hsolu = H1 + H2 + H3
- Where Hsolu may have a positive sign (energy absorbed) or a negative sign
(energy released)
Enthalpy of hydration- the enthalpy changes associated with the disposal of a gaseous
solute in water
Henry’s Law- p=kc
P represents the partial pressure of the gaseous solute above the solution
C represents the concentration of the dissolved gas
K is a constant characteristic of a particular solution
Raoult’s Law- Psolu=XsolventPosolvent
Psolu is the observed vapor pressure of the solution
Xsolvent is the mole fraction of solvent
Posolvent is the vapor pressure of the pure solvent
Colligative properties- freezing point depression, boiling point elevation, and osmotic
pressure
Semipermeable membrane- it allows solvent but not solute molecules to pass through
Concepts:
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Boiling-Point elevation
Freezing-Point depresión
T1=KBM solute KB = molar boiling point elevation constant
T2=KfM solute Kf = molar freezing point depression
M = molality of the solute in the solution (moles solute/ kg solvent)
T1 = bp elevation/ difference between bp of solution and bp of pure solvent
T2 = fp depression/ difference between fp of the solution and fp of the pure solvent
Mass Percent = [mass of solute/ mass of solution] x 100%
Molarity, M
M = [mole of solute/ L of solution]
Molality, m
m = [mole of solute/ kilogram of solvent]
Example Problems
Number 1: A solution is prepared by mixing 1.00g ethanol with 100.0g H2O to give a
final volume of 101mL. Calculate the molarity, mass %, mole fraction, and molality of
ethanol in this solution.
o Molarity:
1.00gC2H5OH X (1 mol C2H5OH)/(46.07g C2H5OH)= 2.17 X 10-² mol C2H5OH
Volume= 101mL X (1L/1000mL)=0.101L
(mol of C2H5OH)/(liters of solution)= (2.17 X 10-2)/(0.101L)=0.215M
o Mass %:
Mole fraction of C2H5OH of C2H5OH=(n C2H5OH)/(nC2H5OH +nH2O)
NH2O= 100.0g H2O X (1mol H2O)/(18.0gH2O)= 5.56 mol
XC2H5OH= (2.17 X 10-2 mol)/(2.17 X 10-2+5.56 mol)=0.00389
o Molality:
(moles of C2H5OH)/(kilogram of H2O)= (2.17 X 10-2)/(100.0g X (1kg/1000g))
=0.217m
Number 2: A certain soft drink is bottled so that a bottle @ 25°C contains CO2 gas at a
pressure of 5.0 atm over the liquid. Assuming that the partial pressute of CO2 in the
atmosphere is 4.0 X 10-4 atm. What is the quilibrium concentration of CO2 before and
after the bottle is opened.
o Henry’s law: CO2 is 32L atm/mol @ 25°C
PCO2=KCO2CCO2
Unopened CCO2 = PCO2/K CO2 = (5.0 atm) / (32. L atm/mol) = 0.16mol/L
Opened C CO2 = PCO2/K CO2 = (4.0 X 10-4 atm)/(32. L atm/mol) = 1.2 X 10-5 mol/L
Number 3: Calculate the expected vapor pressure @ 25°C to a solution prepared by
dissolving 158.0g of common table sugar (molar mass 342.3g/mol) in 643.5 cm3 of
water. At 25°C, the density of water is 0.9971g/cm3 and the vapor pressure is 23.76 torr
o Psoln = XH2OP°H2O
Moles of surcose = (158.0g X (1mol/342.3g)) surcose= 0.4616 mol surcose
643.5 cm3H2O X (0.9971gH2O)/(cm3H2O) = 641.6gH2O
XH2O = (mol H2O)/(mol H2O +mole surcose)= 0.9873
Psoln = XH2OP°H2O = (0.9873)(23.76 torr) = 23.46 torr
Number 4: A solution was prepared by dissolving 18.00g glass in 150.0g water. The
resulting solution was found to have a boiling point of 100.34°C. Calculate the molar
mass of glucose.
o ∆T = Kbmsolute
∆T = 100.34°C –100.00°C = 0.34°C
m solute = ∆T/Kb = (0.34°C)/(0.51°Ckg/mol) = 0.67mol/kg
m solute = 0.67 mol/kg = (mol solute)/(kg solvent) = (n glucose)/(0.1500kg)
n glucose = (0.67 mol/kg)(0.1500kg) = 0.10mol
Number 5: A chemist is trying to identify a unknown hormone which controls
metabolism, by determining its molar mass. A sample weighing 0.546g was dissolved in
15.0g benzene, and the freezing-point depression was determined to be 0.240°C.
Calculate the molar mass of the hormone.
Mhormone = ∆T/kf = (0.240°C)/(5.1206kg/mol) = 4.69 X 10-2 mol/kg
4.69 X 10-2 mol/kg = m solvent = (mol hormone)/(0.0150kg benzene)
mol hormone = 7.04 X 10-4 mol
(0.546g)/(7.04 X 10-4mol) = (Xg)/(1.00mol)
X = 776g/mol