Untitled - Kelly Walsh High School

Solutions 181
Don’t Forget!
Using Raoult’s law:
P solution X solvent P° solvent X solute P° solute
Psolution (0.8858544) (114 torr) (0.1141456) (197 torr)
123.47408 123 torr
In our next example, we will show an example of freezing point depression and
boiling point elevation. This will require us to use the equation ∆T f iK fm. In
this example, we will use a nonelectrolyte, so we will not need the van’t Hoff
factor (or simply i 1).
Determine both the freezing point and boiling point of a solution containing 15.50 g
of naphthalene, C 10H 8, in 0.200 kg of benzene, C 6H 6. Pure benzene freezes at
278.65 K and boils at 353.25 K. K f for benzene is 5.07 K/m and its K b is 2.64 K/m.
We will begin by calculating the change in the freezing point, ∆T f, to answer this
problem. The problem gives us the value of K f, and we are assuming that i = 1.
Therefore, we need to know the molality of the solution to find our answer. To
determine the molality, we will begin by determining the moles of naphthalene.
Naphthalene has a molar mass of 128.17 g/mol. Thus, the number of moles of
naphthalene present is:
15.50 g
0.120933 mol (unrounded)
128.17 g/mol
The molality of the solution, based on the definition of molality, would be:
0.120933 mol
0.200 kg
0.604665 m (unrounded)
We can enter the given values along with the calculated molality into the freezing
point depression equation:
T iKfm (1) (5.07 K/m) (0.604665 m) 3.06565 K 3.07 K
Tf (278.65 3.07) K 275.58 K (2.43C)
You must subtract the T value from the normal freezing point to get the freezing
point of the solution.