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Tro Chemistry - CH12-20


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Most matter we encounter is in the form of ____.
Mixture in which atoms and molecules intermingle on an atomic scale
A solution that draws water to itself
Seawater is an example of a ____ solution.
SOLUTION (specific)
Homogenous mixture of two or more substances.
The majority component of a solution is the ____, and the minority component is the ____.
solvent, solute
What is the solvent and solute in seawater?
SOLVENT: water
Substances tend to combine into uniform mixtures rather than separating into pure substances, unless ________________.
unless it is highly unfavorable energetically.
Why does fluid flow out of body cells and into seawater when it flows through the intestines?
The body cells have less dilute fluid than seawater, and tend toward mixing with it.
Solution where water is the solvent, ex. club soda
What is the solvent and solute in club soda?
SOLVENT: water
SOLUTE: carbon dioxide
Solution where both the solvent and solute are gases
Solution where both the solvent and solute are solids
Brass (copper/zinc) and other alloys are examples of what?
Solid solutions
Amount of a substance that will dissolve in a given amount of solvent
Per 100g H2O, at 25C, what is the solubility of NaCl?
Is grease soluble in water?
Chemical systems tend toward ____ potential energy.
Why do protons and electrons move toward each other?
According to Coulomb's law, potential energy decreases as the separation between particles with opposite charges decreases
If 2 ideal gases that normally would not interact with each other are placed in a split container that then has the barrier removed, they will mix.
TRUE - product of entropy
Measure of energy randomization/dispersal in a system
Why will ideal gases mix if placed in a split container that then has the barrier removed?
Because energy will spread out whenever it is not restricted from doing so
In what 3 places do intermolecular forces occur during solution formation?
1) solvent and solute particles
2) solvent and solvent particles
3) solute and solute particles

What is the first step in solution formation?
Separating the solute into constituent particles; always endothermic because energy is required to overcome solute-solute forces
What is the second step in solution formation?
Separating solvent particles from each other; always endothermic because energy is required to overcome solvent-solvent forces
What is the third step in solution formation?
Mixing solvent and solute particles; always exothermic as energy is released as solute and solvent particles interact
Enthalpy of solution (delta H[soln]) is a sum of enthalpy changes in each step:

delta H[soln] = delta H[solute] + delta H[solvent] - delta H[mix]

If delta H[solute+solvent] = delta H[solvent], delta H[soln] = 0

If delta H[soln] < 0, exothermic

If delta H[soln] > 0, endothermic
Delta H[hydration] = delta H[solvent] + delta H[mix]

* enthalpy change when 1 mol of gaseous solute ions are dissolved in water

Delta H[hydration] is typically _____ for ionic compounds.
In ionic compounds, delta H[solute] = _________.
-delta H[lattice]
What are the three possible absolute values of delta H in ionic aqueous solutions?
1. delta H[solute] < delta H[hydration], exothermic, ex. LiBr, KOH

2. delta H[solute] > delta H[hydration], endothermic, ex. NH4NO3, AgNO3

3. delta H[solute] = delta H[hydration], ex. NaCl, NaF
Describe the dynamic equilibrium attained when a solid solute is dissolved in a liquid solvent.
At first the rate of solute dissolution greatly exceeds the rate of its redeposition, but as its concentration increases in the liquid solvent the rates of dissolution and redeposition become equal (dynamic equilibrium).
Solution in which the dissolved solute has reached dynamic equilibrium with undissolved solute-- if additional solute is added, it will not dissolve
Solution in which there is less than the equilibrium amount of dissolved solute-- if more solute is added, it will dissolve
Unstable solution where more than the equilibrium amount of dissolved solute exists
Adding a solid block of sodium acetate to a solution of sodium acetate causes dramatic crystal-like formations to occur as sodium acetate precipitates out of the solution. What was the solution's characteristic?
The solubility of most solids in water ____ when temperature increases.
A method of purifying solids by exploiting how solubility of solids changes with temperature. Enough of the solid is added to water to create a saturated solution at elevated temperature. Then, as the solution returns to room temperature, it becomes supersaturated and the excess solid precipitates out of the solution as purified crystalline structures.
The solubility of gases in liquids ____ when temperature increases.
Gas solubility in liquid ___ as the pressure of gas above the liquid increases.
Solubility of gases increases with increasing pressure

solubility of gas (M) = Henry's law constant X partial pressure of gas (atm)

S[gas] = k[H] X P[gas]

1.3 X 10^(-3) M/atm
6.1 X 10^(-4) M/atm
3.4 X 10^(-2) M/atm
5.8 X 10^1 M/atm
3.7 X 10^(-4) M/atm
Why does NH3 have a greater kH than O2, N2, CO2 and He?
NH3 is polar and has greater solubility in water
What pressure of CO2 is needed to keep CO2 concentration at 0.12M at 25C?
S[gas] = kH[CO2] X P[CO2]
S[gas] = 0.12M
kH[CO2] = 3.4 X 10^(-2) M/atm (from table)
P[CO2] = ?

.12M = (3.4 X 10^(-2) M/atm) X P[CO2]

P[CO2] = .12M/(3.4 X 10^(-2) M/atm)

P[CO2] = 3.5 atm

What is the solubility of O2 in H2O at 25C, exposed to air at 1.0atm. P[O2] = 0.21atm
S[O2] = ?
kH[O2] = 1.3 X 10^(-3) M/atm (from table)
P[O2] = 0.21atm

S[O2] = (1.3 X 10^(-3) M/atm) X 0.21atm

S[O2] = 0.000273 M

Molarity (mol/L) = amount solute (mol) / volume solution (L)
Molality (mol/kg) = amount solute (mol) / mass solvent (kg)
Mole fraction (no unit) = amount solute (mol) / total amount solute+solvent (mol)
Mole percent (%) = (amount solute (mol) / total amount solute+solvent (mol)) X 100%
Parts by mass (%,ppm,ppb) = (mass solute/mass solution) X multiplication factor

* percent by mass (%) MF = 100
* parts per million by mass (ppm) MF = 10^6
* parts per billion by mass (ppb) = 10^9

Parts by volume (%,ppm,ppb) = (volume solute/volume solution) X multiplication factor

* percent by mass (%) MF = 100
* parts per million by mass (ppm) MF = 10^6
* parts per billion by mass (ppb) = 10^9

Both molarity and molality vary with temperature.
FALSE - only molarity
How many mL of 10.5% sucrose (C12H22O11) solution must you consume to obtain 78.5g of sucrose?
Solution Density = 1.04g/mL
Percent by Mass[sucrose] = 10.5%
D[solution] = 1.04 g/mL
M[sucrose] = 78.5g
mL = ?

Percent by Mass as Ratio = 10.5g sucrose / 100g solution

Density = 1mL/1.04g
78.5g sucrose X (100g solution/10.5g sucrose) X (1mL/1.04g)
= 719 mL solution

How much sucrose is in 355mL of a 11.5% sucrose solution?
D = 1.04 g/mL
V[solution] = 355mL
PBM[sucrose] = 11.5% = 11.5g sucrose / 100g solution
D[solution] = 1.04g/mL
M[sucrose] = ?

355mL solution X (1mL/1.04g) X (11.5g sucrose/100g solution)
= 39.25g sucrose

17.2g C2H6O2 is dissolved in 0.5kg H2O, with a final solution volume of 515mL. Calculate molarity, molality, percent by mass, mole fraction and mole percent for this solution.
M[C2H6O2] = 17.2g
M[H2O] = 0.5kg
V[soln] = 515mL
Molar Mass[C2H6O2] = 62.07g (via periodic table)

mol C6H6O2 = 17.2g C2H6O2 X (1mol C2H6O2/62.07g C2H6O2) = 0.2771 mol C2H6O2

MOLARITY(M) = 0.2771 mol C2H6O2 / 0.515 L soln
= 0.538M

MOLALITY (m) = 0.2771 mol C2H6O2 / 0.5kg H2O
= 0.554

PBM = [17.2g/(17.2g + (5 X 10^2g)] X 100%
= 3.33%

1) (5 X 10^2 g H2O) X (1 mol H2O/18.02g H2O) = 27.75 mol H2O

2) 0.2771 mol C2H6O2 / (0.2771 mol + 27.75 mol)
= 9.89 X 10^(-3)

MOLE PERCENT = (9.89 X 10^(-3)) X 100%
= 0.989%

What is the molarity of a 6.55% by mass glucose solution?
D = 1.03g/mL
PBM[glucose] = 6.55g glucose / 100g soln
Molar Mass[glucose] = 180.16g
D[soln] = 1mL / 1.03g

1) 6.55g glucose X (1 mol glucose / 180.16g glucose) = 0.03636 mol glucose

2) 100g soln X (1mL/1.03g) X ((10^(-3)L)/mL) = 0.09709 L soln

3) 0.03636 mol glucose / 0.09709 L soln = 0.374M glucose

Property of a solution that depends on the number of particles dissolved within it
When 1 mol of nonelectrolyte is dissolved in water, it forms 1 mol of dissolved particles.
When 1 mol of electrolyte is dissolved in water, it forms 1 mol of dissolved particles.
FALSE, forms > 1 mol of dissolved particles
The vapor pressure of a solution is ____ than the vapor pressure of a pure solvent.
Why is the vapor pressure of a solution lower than that of a pure solvent after a nonvolatile solute is added?
1) The solute particles interfere with the solvent particles' ability to vaporize. The rate of condensation rises above the rate of vaporization, and only returns to equilibrium when the concentration of gas phase molecules decreases, lowering the vapor pressure.

2) Tendency toward mixing.

Vapor pressure of solution = mole fraction of solvent X vapor pressure of pure solvent at the same temperature

P[solution] = X[solvent] X P^o[solvent]

A water sample at 25C contains 0.9 mol water and 0.1 mol sucrose. P^o[solvent] = 23.8torr

What is the vapor pressure of the solution?

P[solution] = 0.9mol H2O X 23.8 torr
= 21.4 torr

Since the solvent particles compose 90% of the soln, the P[soln] is 90% of P^o[solvent].

Vapor pressure lowering = vapor pressure of pure solvent - vapor pressure of solution

Delta P = P^o[solvent] - P[soln]

For TWO components:
delta P = X[solute] X P^o[solvent]

* X[solute] = mole fraction of solute

Calculate the vapor pressure of a solution (99.5g sucrose, C12H22O11 + 300mL water, H2O).
T = 25C
D = 1g/mL
P^o[H2O] = 23.8torr

M[sucrose] = 99.5g
Molar Mass[sucrose]= 342.3g
V[H2O] = 300mL
Molar Mass[water] = 18.02g
D[H2O] = 1g/mL
P^o[H2O] = 23.8torr
T = 25C
P[soln] = ?

1a) 99.5g sucrose X (1mol sucrose/342.3g sucrose) = 0.2907mols sucrose

1b) 300mL water X (1g/1mL) X (1mol water/18.02g water) = 16.65mol water

2) X[water] = 16.65mol/(.2907mol+16.65mol) = 0.9828 (mole fraction of solvent)

3) P[soln] = 0.9828 X 23.8 torr
= 23.4 torr

Solution that follows Raoult's Law at all concentrations
What is the Raoult's Law equation for the total vapor pressure of a two-component solution?
P[tot] = P[A] + P[B]

* P[A] = X[A] * P^o[A]
* P[B] = X[B] * P^o[B]

+3.95g carbon disulfide (CS2)
+2.43g acetone (CH3COCH3)
P^o[CS2] = 515torr
P^o[CH3COCH3] = 332torr
Assume ideal behavior.
What is the total vapor pressure of the solution?

M[CS2] = 3.95g
Molar Mass[CS2] = 76.15g
P^o[CS2] = 515torr
M[acetone] = 2.43g
Molar Mass[acetone] = 58.08g
P^o[acetone] = 332torr

1a) 3.95g CS2 X (1mol/76.15g CS2) = 0.05187mol CS2

1b) 2.43g acetone X (1mol/58.0g acetone) = 0.04184mol acetone

2a) X[CS2] = 0.05187mol/(0.05187mol + 0.04184mol) = 0.5535 (mole fraction [CS2])

2b) X[acetone] = 1 - 0.5535 = 0.4465

3a) P[CS2] = 0.5535(515torr) = 285torr

3b) P[acetone] = 0.4465(332torr) = 148torr

4) P[tot] = 285torr + 148torr = 433torr

Solutions have lower melting points than pure solvents
Solutions have higher boiling points than pure solvents
What equation describes how far a solution's freezing point is lowered relative to the pure solvent?
Change in freezing point temperature (degrees C) = molality[soln] (mols solute/kg solvent) X freezing point depression constant[pure solvent]

Delta Tf = m X Kf

Calculate the freezing point of a 1.7m aqueous ethylene glycol solution.
m = 1.7
Kf[ethylene glycol] = 1.86C/m
Freezing Point[water] = 0C

delta Tf = 1.7m X 1.86C/m

Freezing Point = 0C - 3.2C
= -3.2C

What equation describes how far boiling point is elevated for solutions relative to the pure solvent?
change in the boiling point temperature (degrees C) = molality[soln] (mols solute/kg solvent) X boiling point elevation constant[pure solvent] (C/m)

delta Tb = m X Kb

How much ethylene glycol (g) must be added to 1kg water to produce a soln that boils at 105C?
M[eg] = ?
Molar Mass[eg] = 62.07g
Kb[eg] = 0.512C/m
M[water] = 1kg
Boiling Point[soln] = 105C
Boiling Point[water] = 100C
Delta Tb = 5C
m[soln] = ?

m = delta Tb/Kb
m = 5C/.512C/m = 9.77m

1kg water X (9.77mol eg/1kg H2O) X (62.07g eg/1mol eg)
= 6.1 X 10^2g eg

Flow of solvent from lower solute concentration soln to higher solute concentration soln
Membrane that selectively permits some substances through, but not others
Pressure required to stop osmotic flow from low to high solute concentrated solns.

* II = MRT
* osmotic pressure = Molarity X (0.08206L*atm/mol*K) X temperature (degrees K)

II (osmotic pressure) = 2.45torr
T = 25C
5.87mg (unknown) / 10mL soln
What is the molar mass of the unknown protein?

T = 25C = 298K

1) M = II/RT
M = [2.45torr X (1atm/760torr)]/[.08206(L*atm/mol*K) X 298K = 1.318 X 10(-4)M

2) 10mL X (1L/1000mL) X [(1.318 X 10^(-4))/1L] = 1.318 X 10(-6)mol

3) Molar Mass = mass protein/mols protein

MM = (5.87 X 10^(-3)g)/(1.318 X 10^(-6)mol)
= 4.45 X 10^3g/mol

Why is the Td of 0.10M NaCl solution 2x that of 0.10M sucrose solution?
1 mol NaCl dissociates into almost 2 mols of ions in solution (van't Hoff factor)
van't Hoff factor = (moles of particles in soln)/(moles of formula units dissolved)

* Expected factors only occur in very dilute solns (as concentration approaches 0), as IRL dissociation is not complete

How is vapor pressure lowering (delta-P) calculated using van't Hoff factor (i)?
delta-P = i(Xsolute) * P^o(solvent)
How is freezing point depression (Td) calculated using van't Hoff factor (i)?
Td = (i*m) X Kf
How is boiling point elevation (delta-Tb) calculated using van't Hoff factor (i)?
delta-Tb = (i*m) * Kb
How is osmotic pressure (II) calculated using van't Hoff factor?
Which solution has the highest boiling point?:
a) .5M C12H22O11
b) .5M NaCl
c) .5M MgCl2

c) .5M MgCl2
Tf(.05m CaCl aq soln) = -0.27C
What is i?
How does it compare to the predicted i?

m = .05
Tf(CaCl aq soln) = -0.27C
Kf = 1.86C
Expected i = 3 (3 mol ions for each mol CaCl2 that dissolves)

Td = (i*m) * Kf

i = Td/m*Kf
i = .27 / (.05m * 1.86C/m)
i = 2.9

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