Basic Calculation Mola,Molali

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    Calculating Molality - ITo calculate Molality (mol/kg), we need the number of moles of

    solute, and the mass of solvent used to dissolve the solute!

    Normally, we are given the mass of solute, and mass of solvent,

    therefore we calculate the moles of the solute from the mass, then

    use the mass of the solvent to calculate the molality.

    Mass (g) of solute

    Moles of Solute

    Molality (m) of solution

    M (g/mole)

    divide by kg water

    Remember:

    Molality is different from Molarity

    Molality is based on mass, and is

    independent of temperature or

    pressure (unlike molarity)

    Because 1 L of H2O weighs 1 kg,

    molality and molarity of dilute

    aquous solutions are nearly identical

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    Calculating Molality - II

    Problem: Determine the molality and molarity of a solution prepared

    by dissolving 75.0g Ba(NO3)2 (s) in to 374.00g of water at 250C.

    Plan: We convert the quantity of Ba(NO3)2 to moles using the molarmass and then divide by the volume of H2O in liters (using water

    density = 0.99707 g/ml3).

    Solution: molar mass of Ba(NO3)2 = 261.32 g/mol

    moles Ba(NO3)2 = = 0.28700 mole75.0 g

    261.32 g/mol

    molality = = 0.76739 m = 0.767 m0.37400 kg

    0.28700 mole

    molarity - we need the volume of solution, and can assume that

    addition of the salt did not change the total volume.374.00 g H2O

    0.99707 g/ml= 375.099 ml = 0.375099 l

    M = = 0.765 M0.375099 l

    0.28700 mole

    Expressing Concentrations in Parts by Mass

    Problem: Calculate the PPB by mass of Iron in a 1.85 g Iron

    supplement pill that contains 0.0543g of Iron.

    Plan: Convert g Fe to grams and then use Fe/ mass pill and

    multiply by 109 to obtain PPB.

    Solution: 0.0543g Fe = 5.43 x 10 - 8 g Fe

    5.43 x 10- 8

    g Fe1.85 g

    x 109 = 2.94 PPB

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    Expressing Concentrations in Parts by Volume

    Problem: The label on a can of beer (340 ml) indicates 4.5% alcohol

    by volume. What is the volume in liters of alcohol it contains?

    Plan: we know the vol% and the total volume so we use the definition

    to find the volume of alcohol.

    Solution:

    = 15.3 ml alcohol4.5 ml alcohol

    100 ml beer340 ml beerVol Alcohol = x

    Expressing Concentrations in Mole Fraction

    Problem: A sample of alcohol contains 118g of ethanol (C2H5OH), and

    375.0g of water. What are the mole fractions of each?

    Plan: we know the mass and formula of each compound so we convert

    both to moles and apply the definition of mole fraction.

    Solution:

    Moles Ethanol = = 2.744 mol Ethanol118g Ethanol

    43g Ethanol/mol

    Moles Water = = 21.94 mol H2O395g H2O

    18g H2O/mol

    XEthanol = = 0.111172.744

    21.94 +2.744

    XWater = = 0.8888321.94

    21.94 +2.744

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    Converting Concentration Units

    Problem: Commercial concentrated Hydrochloric acid is 11.8 M HCl

    and has a density of 1.190g/ml. Calculate the (a) mass% HCl,

    (b) molality and (c) mole fraction of HCl.Plan: We know Molarity and density. (a) For mass% HCl we need the

    mass of HCl and water (the solvent). Assume 1L of solution, from

    the density we know the mass of the solution, and from the

    molecular mass of HCl we calculate its mass. (b) We know moles

    of HCl and mass of water (c) we use moles HCl from(a) and use

    the mass of water to get moles of water then calculate mole

    fractions and add them to check!

    Solution:

    (a) assume 1L of HCl solution 11.8 moles HCl

    11.8 moles HCl x = 430.228g HCl36.46g HCl

    mole HCl

    Converting Concentration Units - II

    (a) cont.

    1L solution x x = 1190g solution1.190 g soln

    mL soln

    1000 mL

    1 L solution

    mass % HCl = x 100% = 36.1536 % HCl430.228 g HCl

    1190.g solution

    (b) mass of H2O = mass of solution - mass of HCl =

    1190g solution - = 759.772g H2O430.228g HCl

    1190g solution11.8 moles HCl

    0.759772 kg H2O= 15.53 m HCl

    (c) 759.772g H2O

    18.016g H2O/mol H2O= 42.172 mole H2O

    Total moles =

    42.172 + 11.8

    = 53.972=54.0XHCl = = 0.219

    11.8

    54.0XH2O = = 0.781

    42.172

    54.0

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    Definitions

    Saturated - A solution containing the maximum amount of solute that

    will dissolve under a given set of conditions.

    Unsaturated - A solution that contains less than the saturation quantity

    of solute. More solute may be added and it will dissolve.

    Supersaturated - A solution prepared at an elevated temperature and

    then slowly cooled so that more than the usual

    maximum amount of solute remains dissolved.

    Equilibrium in a Saturated Solution

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    For all gasses: Hsolute 0, so Hsoln < 0Gas solubility always decreases with increasing T!

    The Effect of Pressure on Gas Solubility:

    Henrys Law: Sgas = kH x Pgas

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    Henrys Law of Gas Solubility

    Problem: The lowest level of oxygen gas dissolved in water that will

    support life is ~ 1.3 x10-4 mol/L. At the normal atmospheric pressure of

    oxygen is there adequate oxygen to support life in water?

    Plan: We will use Henrys law and the Henrys law constant for oxygen

    in water with the partial pressure of O2 in the air to calculate the amount

    of O2 dissolved in water under atmospheric conditions.

    Solution:

    Soxygen = kH x PO2 = 1.3 x 10-3 mol x (0.21 atm)

    liter atm

    SOxygen = 2.7 x 10- 4 mol O2 / liter

    .

    The Henrys law constant for oxygen in water is 1.3 x 10-3 mol

    liter atm

    and the partial pressure of oxygen gas in the atmosphere is ~21%,

    or 0.21 atm.

    .

    This is adequate to sustain life in water!

    Effect of Temperature: Heat of Solution

    Solubility increases with temperature if the solution process is

    endothermic:

    Solute + Solvent + Heat Solution

    Solubility decreases with temperature if the solution process is

    exothermic:

    Solute + Solvent Solution + Heat

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    Predicting the Effect of Temperatureon Solubility - I

    Problem: From the following information, predict whether the

    solubility of each compound increases or decreases with an increase in

    temperature.

    (a) CsOH Hsoln = -72 kJ/mol

    (b) When CsI dissolves in water the water becomes cold

    (c) KF(s) K+

    (aq) + F-(aq) + 17.7 kJ

    Plan: Write a chemical reaction that includes heat being absorbed(left) or released (right). If heat is on the left, a temperature increase

    shifts to the equilibrium to the right, so more solute dissolves, and visa

    versa.

    Solution:

    (a) The negative H indicates that the reaction is exothermic, so

    when one mole of Cesium Hydroxide dissolves, 72 kJ of heat is released.

    H

    2O

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    Predicting the Effect of Temperature

    on Solubility - II

    (a) continued

    CsOH(s) Cs+

    (aq) + OH-(aq) + Heat

    A higher temperature (more heat) decreases the solubility of CsOH.

    H2O

    (b) When CsI dissolves, the solution becomes cold, so heat is absorbed.

    CsI(s) + Heat Cs+

    (aq) + I-(aq)

    H2O

    A higher temperature increases the solubility of CsI.

    (c) When KF dissolves, heat is on the product side, and is given off

    so the reaction is exothermic.

    KF(s) K+

    (aq) + F-(aq) + Heat

    H2O

    A higher temperature decreases the solubility of KF

    Colligative Properties of Solutions

    17.4) Vapor Pressure Lowering - Raoults Law

    17.5) Boiling Point Elevation

    and Freezing Point Depression

    17.6) Osmotic Pressure

    Well start by focusing on nonvolatile nonelectrolitic solutes

    (e.g. sucrose in H2O)

    Later: volatile solutes, electrolytic solutes

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    See Figure 17.9

    Raoults Law:

    Psolvent = Xsolvent.

    Psolvent

    X is the mole fraction

    Figure 17.8, Zumdahl

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    Example: Vapor Pressure Lowering

    Problem: Calculate the vapor pressure lowering (

    P) when 175g of

    sucrose is dissolved into 350.00 ml of water at 750C. The vapor

    pressure of pure water at 750C is 289.1 mm Hg, and itsdensity is 0.97489 g/ml.

    Plan: Calculate the change in pressure from Raoults law using the

    vapor pressure of pure water at 750C. We calculate the mole

    fraction of sugar in solution using the molecular formula of

    sucrose and density of water at 750C.

    Solution: molar mass of sucrose ( C12H22O11) = 342.30 g/mol

    175g sucrose

    342.30g sucrose/mol= 0.51125 mol sucrose

    350.00 ml H2O x 0.97489g H2O = 341.21g H2O

    ml H2O 341.21 g H2O

    18.02g H2O/mol= 18.935 mol H2O

    Vapor Pressure Lowering (cont)

    Xsucrose =mole sucrose

    moles of water + moles of sucrose

    = = 0.26290.51125 mole sucrose

    18.935 mol H2O + 0.51125 mol sucrose

    P = Xsucrose * PoH2O = 0.2629 x 289.1 mm Hg = 7.600 mm Hg

    Vapor Pressure lowering:

    P = Xsolute. Psolvent

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    Boiling Point Elevation

    Recall: The normal boiling point (Tb) of a liquid is the

    temperature at which its vapor pressure equals atmospheric

    pressure.

    Nonvolatile solutes lower the vapor pressure of a liquid

    greater temperature required to reach boiling point

    Boiling Point Elevation:

    Tb = Kbmsolute

    Kb = Molal boiling point constant for given liquid

    msolute = molal solute concentration

    Kb = Molal boiling point constant for given liquid

    See Table 17.5

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    See Figure 17.3

    Freezing Point

    Depression:

    Tf= Kfmsolute

    Kf= Molal freezing point constant for given liquid

    See Table 17.5

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    See Figure 17.12

    Example: Boiling Point Elevation and Freezing PointDepression in an aqueous solution

    Problem: We add 475g of sucrose (sugar) to 600g of water. What will

    be the Freezing and Boilingpoints of the solution?

    Plan: Find the molality of the sucrose solution, and apply the equations

    for FP depression and BP elevation using the constants from table 17.5.

    Solution: Sucrose (C12H22O11) has molar mass = 342.30 g/mol475g sucrose

    342.30gsucrose/mol= 1.388 mole sucrose

    molality = = 2.313 m1.388 mole sucrose0.600 kg H2O

    Tb = Kb.m = (2.313m)= 1.18oC BP = 100.00oC + 1.18oC

    = 101.18oC

    0.512oC

    m

    Tf= Kf.m = (2.313 m) = 4.30oC

    FP = 0.00oC -4.30oC= -4.30oC

    1.86oC

    m

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    Example: Boiling Point Elevation and Freezing Point

    Depression in a Non-Aqueous Solution

    Problem: Calculate the Boiling Point and Freezing Point of a

    solution having 257g of napthalene (C10H8) dissolved into 500.00g ofchloroform (CHCl3).

    Plan: Just like the first example.

    Solution: napthalene = 128.16g/mol chloroform = 119.37g/mol

    molesnap = =2.0053 mol nap257g nap

    128.16g/mol

    molality = = = 4.01 mmoles nap

    kg(CHCl3)

    2.0053 mol

    0.500 kg

    Tb

    = Kb

    .m = (4.01m) = 14.56oC normal BP = 61.7oC

    new BP = 76.3oC

    3.63oC

    m

    Tf= Kf.m = (4.01m) =18.85oC normal FP = - 63.5oC

    new FP = - 82.4oC

    4.70oC

    m

    Osmosis: The flow of solvent through asemipermeable membrane into a solution

    The semipermeable membrane allows solvent

    molecules to pass, but not solute molecules

    See Figures

    17.15, 17.16

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    Osmotic Pressure, = MRT

    (similar to ideal gas law!)

    R = gas constantM = molar concentration of solute

    Determining Molar Mass fromOsmotic Pressure

    Problem: A physician studying hemoglobin dissolves 21.5mg of the

    protein in water at 5.0oC to make 1.5 ml of solution in order to

    measure its osmotic pressure. At equilibrium, the solution has an

    osmotic pressure of 3.61 torr. What is the molar mass (M) of the

    hemoglobin?

    Plan: We know, R, and T. We convert from torr to atm, and T

    from oC to K, and then use the osmotic pressure equation to solve formolarity (M). Then we calculate the number of moles of hemoglobin

    from the known volume and use the known mass to findM.

    Solution:

    P = 3.61 torr. = 0.00475 atm1 atm

    760 torr

    Temp = 5.00C + 273.15 = 278.15 K

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    Molar Mass from Osmotic Pressure

    M = = = 2.08 x10

    - 4

    M

    RT

    0.00475 atm

    0.082 L atm (278.2 K)mol K

    Finding # moles of solute:

    n = M . V = . 0.00150 L soln = 3.12 x10 - 7 mol2.08 x10 - 4 mol

    L soln

    Calculating molar mass of Hemoglobin (after changing mg to g):

    M = = 6.89 x104 g/mol0.0215 g

    3.12 x10-7 mol

    Concentration from osmotic pressure:

    We started with nonvolatile nonelectrolytes (e.g. sucrose/H2O)

    Now lets look at volatile solutes and electrolytes

    Colligative Properties of Solutions

    17.4) Vapor Pressure Lowering - Raoults Law

    17.5) Boiling Point Elevation

    and Freezing Point Depression

    17.6) Osmotic Pressure

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    Colligative Properties of VolatileNonelectrolyte Solutions

    From Raoults law: Psolvent = Xsolvent. P0solvent and Psolute = Xsolute

    . P0soluteConsider a solution having equal molar quantities of acetone and

    chloroform, Xacetone = XCHCl3 = 0.500. At 350

    C, the vapor pressure ofpure acetone = 345 torr and pure chloroform = 293 torr.

    Determine the vapor pressure of the solution and the partial

    pressure of each component. What are the mole fractions, X, of each

    component in the vapor phase?

    Pacetone = Xacetone. P0acetone = 0.500

    . 345 torr = 172.5 torr

    PCHCl3 = XCHCl3. P0CHCl3 = 0.500

    . 293 torr = 146.5 torr

    From Daltons law of partial pressures we know that XA =PAPTotal

    Xacetone = = = 0.541Pacetone

    PTotal

    172.5 torr

    172.5 + 146.5 torr

    XCHCl3 = = = 0.459PCHCl3PTotal

    146.5 torr

    172.5 + 146.5 torr

    Total Pressure = 319.0 torr

    Vapor is enriched in acetone!

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    Non-ideal solutions Ideal behavior is approached when solute and solvent are involvedin similar intermolecular interactions (Hsoln = 0)

    When weaker solvent-solute interactions occur, heat is removed

    upon dissolving (Hsoln > 0) the observed vapor pressure is higherthan ideal

    When stronger solute-solvent interactions occur, heat is releasedupon dissolving (Hsoln < 0)

    the observed vapor pressure is lowerthan ideal

    Figure 17.11

    Colligative Properties of Ionic Solutions

    For ionic solutions we must take into account the number of ions present!

    i = vant Hoff factor = ionic strength, or the number of ions present

    For Electrolyte Solutions:

    vapor pressure lowering: P = iXsoluteP0

    solvent

    boiling point elevation: Tb = iKb m

    freezing point depression: Tf= iKf m

    osmotic pressure: = iMRT

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    Non-ideal Behavior

    of Electrolyte

    Solutions

    (0.05m aqueous)

    See Table 17.6

    Non-ideal

    Behavior of

    Electrolyte

    Solutions

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    Gallery Section (p. 508)

    Plane de-icing and car antifreezing-

    ethylene glycol (C2H6O2)

    Biological antifreeze in fish- glycerol (C2H8O3)

    Salts for de-icing streets- CaCl2

    Water volume regulation in cells-

    extracellular Na+ and osmosis

    (also, salt cured foods!)

    Zone refinement- purification of crystals

    Key theme in each: the inability of solutes to cross a

    phase barrier changes the properties of the solvent.

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    The structure and function of a soap

    Micelles- stabilize nonpolar colloids in polar solvents