Changing melting and boiling pointsChanging melting and boiling points
Putting salt on sidewalks and roads in the winter helps because it lowers the freezing point of water by a few degrees.
As a result, the roads won’t be icy unless the temperature falls below the new, lower freezing point (or unless there is too much snow or ice for the salt to melt).
How does this work?
Take a look at ice (at oC)
The ice/water mixture is right at its freezing point, which means some molecules are freezing, and others are melting.
Adding solute particles makes it more difficult for the water molecules to come together to form crystals.
Adding solute particles also interferes with boiling.
This happens because the dissolved ions/molecules make it more difficult for the solvent particles (usually water) to escape.
In this picture imagine that the green particles are water molecules, and the red are dissolved ions.
The ions simply block the water molecules from leaving the liquid state.
Calculating the expected amount of freezing pt. depression or boiling point elevation:
Tf = Kf (m) (# of particles)
Tf = decrease in freezing point
Kf = freezing point constant of the
solvent (usually water)
m = molality = moles of solute ÷ kg of solvent
# of particles = the number of dissolved ions per formula unit of the dissolved compound. For example: NaCl dissolves to create 2 particles/ions. CaCl2 dissolves to create 3, etc.
Molecular substances (like sugar) = 1 particle, since the molecule itself does not break down during the dissolving process.
What would be the freezing point if 125 g of NaCl is added to 750 g of water?
Tf = Kf (m) (# of particles)
Kf of water is 1.86 oC/m
The molality of this solution is 2.85 m (125 g of NaCl = 2.14 mol, then divide by 0.750 kg)
# of particles created by dissolving NaCl = 2
Answer = -10.6 oC (the original f.p. for water is 0 - Tf )
How many grams of sugar (C12H22O11) are added to 150 g of water for a f.p. = -3.5ºC?
Ans =97 g