Untitled - Kelly Walsh High School

Solutions 183
Quick Tip
A molar mass (molecular weight) less than 1.0 g/mol indicates an error. No
molecular weight is less than the atomic weight of hydrogen.
In the preceding examples, we saw how to deal with nonelectrolytes. If the solution
contains an electrolyte, there will only be one change necessary. This
change will be to enter the value of the van’t Hoff factor. We will see how to do
this in the next example.
Determine both the freezing point and boiling point of a solution containing
15.50 g of sodium sulfate, Na 2SO 4, in 0.200 kg of water. Pure water freezes at
0.00°C and boils at 100.00°C. K f for water is 1.86°C/m and its K b is 0.52°C/m.
We will follow the same procedure as in the naphthalene/benzene example
above. You may wish to look over these examples in parallel to see exactly
where the difference between an electrolyte and nonelectrolyte manifests itself.
We will again begin by calculating the freezing point, ∆T f. The problem gives us
the value of K f. In solution, the strong electrolyte, sodium sulfate, ionizes as:
Na 2SO 4(aq) l 2 Na (aq) SO 4 2 (aq)
From this relationship, we can see that each Na 2SO 4 produces three ions. The
production of three ions means that the van’t Hoff factor, i, is 3. 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 sodium sulfate. Sodium sulfate has
a molar mass of 142.04 g/mol. Thus, the number of moles of sodium sulfate
present is:
15.50 g
0.10912419 mol (unrounded)
142.04 g/mol
The molality of the solution, based on the definition of molality, would be:
0.10912419 mol
0.200 kg
0.545620951 m (unrounded)
We can enter the given values along with the calculated molality into the freezing
point depression equation:
T iKfm (3) (1.86C/m) (0.545620951 m) 3.0445649C 3.04C
Tf (0.00 3.04)°C 3.04°C