Stoichiometry
Pure water freezes at 0C (32F)
If 10 grams (0.35 ounces) of sodium chloride (table salt) is
dissolved in 100 grams (3.53 ounces) of water, the freezing point
goes down to 5.9C (21.4F).
Why does the freezing point of water drop when sodium chloride
is added to it?
When the rate of freezing is the same as the rate of melting,
the amount of ice and the amount of water won't change.
The ice and water are said to be in dynamic equilibrium with
each other.Here's the same container with the water at 0C, only
this time the water contains salt molecules. Adding salt disrupts
the equilibrium.
The salt molecules dissolve, but do not attach easily to the
solid ice. There are fewer water molecules; some of the water has
been replaced by salt. The number of water molecules able to be
captured by the ice decreases, so the rate of freezing slows. The
rate of melting of the ice is unchanged by the salt so melting is
now occurring faster than freezing. So the ice eventually melts.
The temperature drops so that the rate of freezing can increase to
reach equilibrium.
In this example, the new freezing/melting point is reached at
-4C. The higher the concentration of salt, the lower the freezing
point drops. If sucrose (table sugar) is used instead of sodium
chloride, will the freezing point of the solution be lower or
higher than that of the salt solution? 10 grams (0.35 ounces) of
sucrose C12H 22O11 in 100 grams (3.53 ounces) of water gives a
solution with a freezing point of 0.56C (31F).Why is the freezing
point of the salt solution lower than that of the sugar
solution?
The reason that the salt solution has a lower freezing point
than the sugar solution is that there are more particles in 10
grams of sodium chloride than in 10 grams of sucrose.
Freezing point depression is a colligative property which is a
property of solutions that depends on the number of dissolved
particles in solution, not on the identities of the solutes. Each
sodium chloride unit comes apart into two ions (a sodium cation and
a chloride anion ) when dissolved in water.
Calculate the change in freezing point ( T f ) relative to the
pure solvent using the equation: T f = i K f mwhere K f is the
freezing point depression constant for the solvent (1.86Ckg/mol for
water)m is the number of moles of solute in solution per kilogram
of solventand i is the number of ions present per formula unit
(e.g., i = 2 for NaCl).
1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of
benzene.
The freezing point of pure benzene is 5.5 oC, and the freezing
point of the mixture is 2.8 oC.
What is the molal freezing point depression constant, Kf of
benzene?Calculating Freezing Point DepressionStep 1: Calculate the
freezing point depression of benzene.Tf = (Freezing point of pure
solvent) - (Freezing point of solution)(5.5 oC) - (2.8 oC) = 2.7
oCStep 2 : Calculate the molal concentration of the
solution.molality = moles of solute / kg of solventmoles of
naphthalene = (1.60 g) (1 mol / 128 g) = 0.0125 mol
naphthalenemolality of solution = (0.0125 mol) / (0.0200 kg) =
0.625 mStep 3: Calculate Kf of the solution.Tf = (Kf) (m)(2.7 oC) =
(Kf) (0.625 m)Kf = 4.3 oC/m
Everyday uses of Freezing Point DepressionWhat ideas do you
have?Salt melting iceCar antifreezeMaking ice cream This is a lab
activity we will be doing in class!
ReferencesRead more: Colligative Properties - Chemistry
Encyclopedia - water, uses, examples, gas, number, equation, salt,
property, Freezing Point Depression, Boiling Point Elevation, Vapor
Pressure Lowering, Osmotic Pressure
http://www.chemistryexplained.com/Ce-Co/Colligative-Properties.html#ixzz1ESZg9En9http://www.worsleyschool.net/science/files/saltandfreezing/ofwater.htmlhttp://chemed.chem.purdue.edu/genchem/probsolv/colligative/kf1.3.html
Image:
http://faculty.clintoncc.suny.edu/faculty/michael.gregory/files/bio%20101/bio%20101%20lectures/chemistry/chemistr.htm
FreezeJordin SparksJ Sparks/Ghost/T. E. Hermansen/Ian
Dench/Mikkel S. EriksenJordin Sparks2007iTunes 10.1.1, QuickTime
7.6.6
