Transcript Salty Bubbles

Salty Bubbles
Presentation by
Sana Panjwani & Alice Yang
Purpose
To determine the influence of
the amount of solute on the
boiling point
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Our materials
We tested 3 different amounts of
NaCl:
15 g
30 g
43g
We kept a constant of 50 mL H2O
each trial
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Time for
Notes
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Key Terms
#2a solute: the substance that will be
dissolved. Ex: NaCl(s)
#2b solvent: substance that will do the
dissolving. Ex: H2O(l)
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#4
Polar dissolve other polars
Ex: H2O and NaCl
Nonpolar dissolves other nonpolars
Motor oil and gasoline
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#15a Dissociation- separation of ions that
occurs when an ionic compound dissolves
NaCl(s)
H2O
Na+ (aq) + Cl- (aq)
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#10 Electrolyte- any substance that
dissociates into ions when dissolved in a
suitable medium or melted. forms a
conductor of electricity.
NaCl is a strong electrolyte, meaning that
most of its bonds will break to form ions.
As opposed to a weak electrolyte and its
few bonds that break into ions.
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 #12 molality (m): number of mole of solute
amount of solvent (kg)
 NOT the same as #2d or #11 molarity (M)
M=
Number of moles of solute
Amount of solvent (L)
m=
Number of mole of solute
Amount of solvent (kg)
 the mass of the solute will not change with varying
temperatures, while the volume of the solvent will.
Therefore, for our experiment, molality would provide a
constant mass to work with.
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#16 Colligative properties- any property of
a solution that is changed by the addition
of a solute.
Ex: adding NaCl to water increases the
boiling point.
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#17b dealing with colligative properties:
Electrolytes in solutions. As stated before,
NaCl, an electrolyte, increases boiling point
when it is added to water.
How to see this mathematically requires the use
of a formula.
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Δt =
Moles of ions
mKb( Per electrolyte in solution )
Δ delta (change)
t temperature
m molality
Kb constant. For water, it is .51 C/m
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Δt =
Δt: we are solving for the change in temp.
to see if increasing amounts of NaCl will
increase the boiling point.
Our first amount of NaCl that we tested
was 15g
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m
molality:
number of mole of solute
amount of solvent (kg)
 Step 1: find number of mole of solute. Use
dimensional analysis
15 g NaCl 1 mol NaCl
58.443 g NaCl
.257 mol NaCl
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m cont.
 Step 2: find amount of solvent (kg)
* remember that 1 mL of water = 1 g of water
50 mL H2O = 50 g H2O = .050 kg H2O
 Step 3: plug back in to find m
m
.257 mol NaCl
5.140 mol/kg
.050 kg H2O
15
Kb
Kb of water is .51 C/m
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Moles of ions
Per electrolyte in solution
 How many moles of ions are there for every
electrolyte? (think back to dissociation)
NaCl(s)
H2O
Na+ (aq)
+ Cl- (aq)
From 1 mol NaCl(s), the reaction yields 1 mol Na
(aq) and 1 mol Cl (aq).
So for every electrolyte in the solution, you have 2
mols of ions.
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Plugging the numbers in.
 Δt = mKb(
Moles of ions
)
Per electrolyte in solution
 Δt= (5.140)(.51)(2)
 Δt= 5.243
we should find that the boiling point of the
solution increases by about 5°C when 15 g
NaCl is added
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Your Turn!
Attempt to find the change in
temperature for an added 30 g NaCl.
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Your turn!
Remember the formulas:
Δt = mKb(
m=
Moles of ions
Per electrolyte in solution
)
Number of moles of solute
Amount of solvent (kg)
Kb= .51 C/m
moles of ions
Per electrolyte in solution
2
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30 g NaCl 1 mol NaCl
58.443 g NaCl
.513 mol NaCl
.050 kg H2O
.513 mol NaCl
10.260
mol/kg
Kb= .51 C/m
moles of ions
Per electrolyte in solution
2
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Δt = mKb (
Moles of ions
Per electrolyte in solution
)
Δt = (10.266) (.51) (2)
Δt = 10.471
The new boiling point temperature should
increase by about 10ºC.
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With 43g NaCl
Δt = mKb(
Moles of ions
Per electrolyte in solution
)
Δt = (14.715)(.51)(2)
Δt =15.009
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Conclusion
An increased amount of NaCl solute in the
tap water will result in an increase in the
boiling point of the solution.
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