Untitled - Kelly Walsh High School

182 CHEMISTRY FOR THE UTTERLY CONFUSED
To calculate the boiling point of the solution we use the relationship:
T b iK bm
We already have the van’t Hoff factor, the K b, and solution molality so we can
simply substitute:
Tb iKbm (1) (2.64 K/m) (0.604665 m) 1.59632 K (unrounded)
Tb (353.25 1.59632) K 354.85 K (81.70°C)
The freezing point depression and boiling point elevation techniques are useful
in calculating the molar mass of a solute or its van’t Hoff factor. In these cases,
you will begin with the answer (the freezing point depression or the boiling
point elevation), and follow the same steps as above in reverse order.
In the next example, we will examine the colligative property of osmotic pressure.
This will require us to use the relationship p i(nRT/V).
A solution prepared by dissolving 7.95 mg of a gene fragment in 25.0 mL of water
has an osmotic pressure of 0.295 torr at 25.0°C. Assuming the fragment is a nonelectrolyte;
determine the molar mass of the gene fragment.
In this example, we need to determine the molar mass (g/mol) of the gene fragment.
This requires two pieces of information—the mass of the substance and
the number of moles. We know the mass (7.95 mg), thus we need to determine
the number of moles present. We will rearrange the osmotic pressure relationship
to n V/RT. We know the solute is a nonelectrolyte so i 1. We can
now enter the given values into the rearranged equation and perform a pressure
and a volume conversion:
(0.295 torr) (25.0 mL) 1 atm
a
L atm
760 torr
a0.0821 b (298.2 K)
mol K ba
1L
1000 mL b 3.96 107 mol
The final step in the problem is to combine the given mass and the moles we
found to give the molar mass. This will require the conversion of the milligrams
given into grams.
(7.95 mg)
(3.96 10 7 mol) a103 g
1mg b 2.01 104 g/mol