15. SOLUTIONS ( Samerah C. Macabaas).pptx

REPORTER:
SAMERAH C. MACABAAS-SULTAN


 Mixture: is a blend of two or more pure substances
that are not chemically combined, but remain the
same individual substances; can be separated by
physical means.
 Two types:
 Heterogeneous
 Homogeneous
REVIEW


 “Hetero” means “different”
 Consists of visibly different substances or phases
(solid, liquid, gas)
 Can be separated by filtering
 Example:
Heterogeneous Mixture


 “Homo” means the same
 has the same uniform appearance and composition
throughout; maintain one phase (solid, liquid, gas)
 Commonly referred to as solutions
 Example:
Homogeneous Mixture
Salt Water


 Solution: is a homogeneous mixture of two or more
substances in a single physical state.
 can be physically separated
 composed of solutes and solvents
Solution
the substance being dissolved
the substance that dissolves the solute
Iced Tea Mix
(solute)
Water
(solvent)
Iced Tea
(solution)
Salt water is
considered a
solution. How can it
be physically
separated?


 The solvent is the largest part of the solution and the
solute is the smallest part of the solution
Solution
S O L V E N T
S O L U T E


1. Gaseous solutions
 it is formed when two or more gases combined.
 The solute is gas and the solvent is gas.
 the solution is in gaseous state.
 Examples: air
Types of Solutions


2. Liquid solutions
 it is a combination of a solvent which is liquid and
the solute that can be a solid, liquid or gas.
 the overall appearance of the solution is in liquid
state.
 if the liquid solution contains solvent as water it is
specifically called as “aqueous solution”.
 Examples: Salt water, ethyl alcohol, and soft drinks
Types of Solutions


3. Solid solutions
 it is formed when two or more solid substances
combined.
 the solute is solid and the solvent is solid.
 the overall appearance of the solution is solid in
phase.
 Examples: alloys, brass, etc
Types of Solutions


Concentration of solution is the amount of solute
dissolved in a solvent at a given temperature
Concentration of solutions
•described as dilute if it has
a low concentration of solute
dissolved
•described as concentrated
if it has a high concentration
of solute dissolved


Molarity is the concentration of a solution expressed
in moles of solute per Liter of solution.
Molarity is a conversion factor for calculations
Molarity (M) = moles of solute
Liters of solution
Molarity


Example 1: What is the molarity of a solution that has
2.3 moles of sodium chloride in 0.45 liters of
solution?
Molarity
2.3 moles NaCl = 5.1M NaCl
0.45 L


Example 2: How many moles of Na2CO3 are there in
10.0 L of 2.0 M solution?
Step 1. derived the formula.
Mol solute=(Molarity) (L solution)
Step 2. apply the new derived formula
Molarity
Mole of solute = 2.0M x 10.0 L
Mole of solute = 20.0 moles Na2 CO3


Example 3: How many moles of KNO3 are needed to
make 450. mL of 1.5 molar solution?
Step 1. convert 450 ml to L of solution
Step 2. derived the formula
mole of solute = Molarity x L of solution
Molarity
Mole of solute = 1.5 x .45L =0.675moles KNO3
450ml x 1L = 0. 45L
1000ml


Example 4: How many grams of NaCl are needed to
make 3.0 L of 1.5 M solution?
Step 1. solve for the mole of solution.
Step 2. convert the mole of solution to gram.
Molarity
1.5 M = mole of solute = 4.5 mole of NaCl
3.0L
4.5 mole of NaCl x 58.44 g of NaCl =262.98 g NaCl
1mole of NaCl


Example 5: How many L of 4.0 M solution can be made
with 132g of NaCl ?
Step 1. convert 132g of NaCl to mole of NaCl.
Step 2. derived the formula since we are looking for the
L of solution. L of solution = mol of solute/ molarity
Molarity
132g of NaCl x 1mole of NaCl= 2.259 mole of NaCl
58.44 g of NaCl
L solution= 2. 259 mole of NaCl = 0.567 L of NaCl
4.0 M


Molality is a measure of concentration of solution
that refers to the number of moles of solute dissolves
in each kilogram of solvent.
Molality (m) = moles of solute
Kilogram of solvent
Molality


Example 1 an 18.0 grams of C6H12O6 is dissolved in 1 Kg of
water. What is the molality of the resulting solution.
Step 1. convert 18.0 g C6H12O6
Step 2. solve for the molality.

Molality
18.0 g of C6H12O6 x 1 mole of C6H12O6=0.1 mole of C6H12O6
180 g of C6 H12
m = 0.1 mol of C6H12O6 = 0.1 Mol of C6H12O6 /Kg of H2O
1kg of H2O


Examples 2. chlorine is often added to water in swimming
pools and drinking to keep the water clear and free of living
organisms. What is the molality of a solution made up of 16.1
g of chlorine in 5000g of water?
Step 1. convert 16.1 g of Cl2 to mole of Cl2
Step 2. convert 5000g of H2 O to Kg of H2 O .
Step 3. solve for the molality
Molality
5000g of H2 O x 1 kg of H2 O = 5Kg of H2 O
1000g of H2 O
Molality = 0.227 mole of chlorine = 0.045 mol of chlorine/kg of water
5kg of water
16. 1g of Chlorine x 1 mole of chlorine = 0.227 mole of chlorine
70.88 g of chlorine


Mole fraction is a measure of concentration of solution
that refers to the number of moles of one component
divided by the total number of moles in a solution.
The component can be solute or solvent.
Mole fraction = mole of component
total mole of solution
Mole Fraction


FORMULA
X solute = mole of solute
X solvent = mole of solvent
Mole Fraction


1. What is the mole fraction of sulfur dioxide (SO2) in an
industrial exhaust gas containing 128.0 g of sulfur
dioxide dissolved in every 1500g of CO2.
Step 1. convert 128.0 gram of SO2 to mol of SO2
Step 2. convert 1500g of CO2 to mol of CO2.
Mole Fraction
Mol SO2 = 128.0 g SO2 x 1 mol SO2 = 1.999 mol SO2
64.04 g SO2
Mol CO2 = 1500g CO2 x 1 mol CO2 = 34. 09 mol CO2
43.99 g CO2


XSO2 = mole of SO2 o
= 1.999 mol SO2 .
1.999 mol SO2 + 34.09 mol CO2
= 0. 05539
Mole Fraction


X = mole of N2 O
= 1. 146 mol N2O
1.146 mol N2O + 31.98 mol O2
= 0. 36
Mole Fraction


Saturation of solutions
•Unsaturated - has a less than the
maximum concentration of solute dissolved
•Saturated - has the maximum
concentration of solute dissolved (can see
solid in bottom of solution)
•Supersaturated -contains greater amount
of solute than that needed to form
saturated solution. (usually requires an
increase in temperature followed by
cooling)


The amount of solute that
dissolves in a certain amount of a
solvent at a given temperature and
pressure to produce a saturated
solution.
Solubility


Nature of solute and solvent
Temperature
pressure
Factors affecting solubility of solids

Nature of solute and solvent
 Chemists use the saying
“like dissolves like”:
 Polar solutes tend to
dissolve in polar
solvents.
 Nonpolar solutes tend to
dissolve in nonpolar
solvents.
Oil is nonpolar while water is
polar. They are immiscible.

Temperature
Generally, the
solubility of
solid solutes in
liquid solvents
increases with
increasing
temperature.
Solubility Curves

To read the
graph, find the
line for the
substance. The
amount that
dissolves at a
given
temperature is
on the y- axis.

How much KNO3
dissolveS in 100g (or
100mL H2O at 50oC?
1.Find the line (green)
2.Find the temperature
and follow up to the
line.(red arrow)
3. Read across to the y-
axis and this is the
answer. (blue arrow)
4. Since it is more than
½-way between 80 and
90, it is 87.


 A point on the line
is a saturated
solution.
 Above the line is
supersaturated.
 Below the line is
unsaturated.

Using Solubility Curves
1. What is the solubility of NaNO3
in 100 g of H2O at 0°C?
2. How many grams of KNO3 will
dissolve in 200g of H2O at 45°C?
3. How much water is needed to
dissolve 190g of NaNO3 at 30°C?
73g NaNO3
75g = ?
100g H2O 200g H2O
150 g KNO3
95g = 190g
100g H2O ? g H2O
200 g H2O

Factors affecting the rate of dissolution
Temperaturee
increased temperature causes
solids to dissolve faster
Shaking
Note: Increasing the amount of solute
DOES NOT increase the rate of dissolving
Shaking (agitation) causes
solids to dissolve faster
Smaller particles dissolve
Faster because they have
more surface area
Particle Sizee


 Miscible liquids can easily dissolve in one another.
Example alcohol and water.
 Immiscible liquids are not soluble in each other.
Example oil and water.


 Colligative property- depends on the concentration
of solute particles but is independent of their nature.
 Types of colligative properties
1. Vapor pressure reduction
2. Boiling point elevation
3. Freezing point depression
4. Osmotic pressure
Colligative properties


 When non-volatile solute is dissolved in solvent, the
vapor pressure of solvent is lowered.
 Solvent molecules on the surface which can escape into
vapor is replaced by solute molecules have little vapor
pressure.


 Since the rate of condensation is faster than the
evaporation the vapor pressure will be reduced and
it is known as vapor pressure reduction.
 The vapor pressure is directly proportional with the
mole fraction of solute.


Formula:
PA = XA P0
A
P0
A – PA = XB P0
A


Example: calculate the lowering of vapor pressure and
the vapor pressure of solution containing 50g
dextrose in 1000g of water. The vapor pressure of
water = 17.535 mm/Hg.
Step 1. convert 50g of C6 H12 O6 to mol of C6 H12 O6
Step 2. convert 1000g of H2 O.
50 g of C6H12O6 x 1 mole of C6H12O6=0.278 mole of C6H12O
180 g of C6 H12
1000 g of H2Ox 1 mole of H2O=55.5 mole of H2O
18 g of H2O


Step 3. solve for the mole fraction of solute.
XB = nB /(nA + nB )
Ste 4. calculate the vapor pressure lowering.
P0
A – PA = XB P0
A
Step 5. find the vapor pressure of solution.
XB= 0.278 = 0.00498
0.278+55.5
P0
A – PA = 0.00498 x 17.535 = 0.0873mm
PA = 17.535 - 0.0873mm = 17.448mm


 The boiling point of a liquid is defined as the
temperature at which the vapor pressure of that
liquid is equal to the atmospheric pressure which is
760 mmHg.
The vapor pressure of the
solution is lower at any given
temperature.
Therefore, a higher temperature
is required to boil the solution
than the pure solvent.


If we represent the difference in boiling point between the pure
solvent and a solution as Tb ,
we can calculate the change in boiling point from the following
formula.
Where:
Tb :boiling point elevation
m : molality (because molality is temperature independent)
Kb : boiling point elevation constant that depends on a particular
solvent used.
Tb =Kb m


Molal boiling point elevation constants
Solvent Kb (C0 /M)
Acetic acid (C2H4O2) 2.93
Benzene (C6H6) 2.67
Carbon tetrachloride (CCL4) 5.02
Chloroform (CHCl3) 3.85
Ethanol (C2H6O) 1.20
Water (H2O) 0.52
Molal boiling point elevation constants


example 1. water with sugar added to it will boil at a
higher temperature than pure water. By how much
will the boiling point of water be elevated if 100g of
sucrose C12H22O11 is added to 500g of water? For
water K is 0.52C0 /m
Step 1. convert 100g of C12H22O11 to mole of C12H22O11
Step 2. convert 500g of water to kg of water.
Tb =Kb m
100g C12H22O11 x 1 mol C12H22O11
342.3 g C12H22O11 = 0.292 mol C12H22O11
500g H2O x 1 kg H2O
1000g H2O = 0.5kg H2O


Step 3. compute the molality of solution
m= mol of solute/ kg of solvent.
Step 4. solve for boiling point elevation of the solution.
Therefore: the boiling point of solution is 0.30 C0 higher than the
boiling point of water thus the boiling point of the solution is
100.30 C0 .
Tb =Kb m
m= 0.292 mol C12H22O11
0.5kg H2O = 0.584 m
Tb = 0.52C0/m x 0.584m
= 0.30 C0


Example 2. what is the boiling point elevation when
12.0g of Iodine is dissolved in 100g of carbon
tetrachloride (CCl4)? K for carbon tetrachloride is
5.02 C0 /m.
Step 1. convert 12.0g of I2 to mole of I2 .
Step 2. convert 100g of CCl4 to kg CCl4?
Tb =Kb m
12.0g I2 x 1 mol I2
254 g I2 = 0.0472 mol I2
100g CCl4 x 1 kg CCl4
1000g CCl4 = 0.1 kg CCl4


Step 4. solve for boiling point elevation of the solution.
Tb =Kb m
m= 0.0472 mol I2
0.1kg CCl4 = 0.472 m
Tb = 5.02 C0/m x 0.47m
= 2.37 C0


 In order for a liquid to freeze it must achieve a very
orderly state that results in the formation of crystals.
 If there are impurities in the liquid, i.e. solutes, the
liquid is inherently less ordered. Therefore, a solution is
more difficult to freeze than the pure solvent so a lower
temperature is required to freeze the liquid.
 normal freezing or melting point
is the temp. at which solid and
liquid are in equilibrium under 1
atm.
 Addition of solute will decrease
the vapor pressure and so will
decrease the freezing point.


 like boiling point elevation, the decrease in freezing
point ( Tf ) is directly proportional to the molality
of the solute. And it is given by the equation
Tf =Kf m
Where:
Tf : freezing point depression
Kf : molal depression constant that depends on the
particular solvent being used
m : molality
Tf =Kf m


Molal freezing point depression constants
solvent Kf (C0 /M)
Acetic acid (C2H4O2) 3.90
Benzene (C6H6) 5.12
naphthalene (C10H8) 7.00
Chloroform (CHCl3) 4.68
Camphor (C10H16O) 40.0
Water (H2O) 1.86
Molal freezing point depression constants


Example 1. while antifreeze protects a car from freezing. It
also protects the car from overheating. Calculate the
freezing point depression of a solution of 100g of ethylene
glycol (C2H6O2) antifreeze in 0.500kg of water . K for water
is 1.86 C0 /m.
Step 1. convert 100g of C2H6O2 to mole of C2H6O2.
Tf =Kf m
100g C2H6O2 x 1 mol C2H6O2
62 g C2H6O2 = 1. 613 mol C2H6O2


Step 3. calculate the freezing point depression.
Tf =Kf m
Addition of antifreeze lower the freezing point of water
by 6.01 C0
Tf =Kf m
m= 1.613 mol C2H6O2
0.5 kg H2O = 3.23 m
Tf = 1.86 C0 /m x 3.23 m = 6.01 C0


Diffusion in liquids:
 Substance tent to move or diffuse from regions of higher
concentration to region of lower concentration so the
differences in concentration disappear.
 By placing water solvent on conc. aqueous salt solution. Salt
moves into water layer and water moves to salt solution till the
new solution becomes uniform.
4. Osmotic pressure


 Osmosis:
on separating conc. Solution from water by semi permeable
membrane ( allow passage of solvent & prevent solute) water
will diffuse from solution of low solute conc. To solution of
higher solute conc.
 The diffusion of solvent through semi permeable membrane is
called osmosis.
4. Osmotic pressure


What is osmotic pressure?
 is the external pressure that must be applied to the
solution in order to prevent it being diluted by the
entry of solvent via osmosis.
Osmotic pressure & concentration
Twice concentration twice osmotic pressure
Osmotic pressure & number of molecules
Osmotic pressure of 2 solutions having the same molal
concentration are identical.
4. Osmotic pressure


Example:
Solutions contains 34.2 g of sucrose in 1000g of water
has the same osmotic pressure as dextrose solution
contains 18 g / 1000g of water.
 No. of moles of sucrose = 34.2/ 342 = 0.1 molal
 No. of moles of dextrose= 18/180 = 0.1 molal
So that the 2 solutions are iso- osmotic
4. Osmotic pressure


Example
what is the osmotic pressure of 1g of sucrose (mol
wt=342) dissolved in 100ml water at 25 degree
celsius? R= 0.082
Step 1. convert 1g of C12 H22 O11 to mol of C12 H22 O11
4. Osmotic pressure
1g C12H22O11 x 1 mol C12H22O11
342 g C12H22O11
=0. 0029 mol C12H22O11


Step 2. solve for the molarity of the solution
M= 0.0029 mol of sucrose/ .1 L = 0.029 M
Step 3. solve for the osmotic pressure.
Л = cRT
= 0.029 x 0.082 x 298
=0.7 atm
4. Osmotic pressure


Colligative properties of solutions provide
a useful means of determining the molar
mass of the unknown substance.
Example:
suppose a 10.0 gram sample of an
unknown compound is dissolved in
0.100 kg of water. The boiling point of
the solution is elevated to 0.433 degree
Celsius above the normal boiling point
of water. What is the molar mass of the
unknown sample?
Determining the molar mass

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