The Foundations of Chemistry

Solution

From Table 14-2, the freezing point of pure benzene is 5.48°C and Kfis 5.12°C/m.

Tf5.48°C4.92°C0.56°C

m0.11 m

The molality is the number of moles of solute per kilogram of benzene, so the number of moles
of solute in 50.0 g (0.0500 kg) of benzene can be calculated.

0.11 m

_?_ mol solute(0.11 m)(0.0500 kg)0.0055 mol solute

mass of 1.0 mol2.2 102 g/mol

molecular weight 2.2 102 amu

You should now work Exercise 61.

EXAMPLE 14-11 Molecular Weight from a Colligative Property

Either camphor (C 10 H 16 O, molecular weight152 g/mol) or naphthalene (C 10 H 8 , molecular
weight128 g/mol) can be used to make mothballs. A 5.2-gram sample of mothballs was
dissolved in 100. grams of ethyl alcohol, and the resulting solution had a boiling point of
78.90°C. Were the mothballs made of camphor or naphthalene? Pure ethyl alcohol has a boiling
point of 78.41°C; its Kb1.22°C/m.

Plan

We can distinguish between the two possibilities by determining the molecular weight of the
unknown solute. We do this by the method shown in Example 14-10, except that now we use
the observed boiling point data.

Solution

The observed boiling point elevation is

TbTb(solution)Tb(solvent)(78.9078.41)°C0.49°C

Using Tb0.49°C and Kb1.22°C/m,we can find the molality of the solution.

molality0.40 m

The number of moles of solute in the 100. g (0.1000 kg) of solvent used is

0.40 (0.100 kg solvent)0.040 mol solute

The molecular weight of the solute is its mass divided by the number of moles.

130 g/mol

The value 130 g/mol for the molecular weight indicates that naphthalene was used to make
these mothballs.

You should now work Exercise 59.

5.2 g



0.040 mol

_?_g



mol

mol solute



kg solvent

0.49°C



1.22°C/m

Tb



Kb

1.20 g solute



0.0055 mol solute

no. of g solute



no. of mol solute

_?_ mol solute



0.0500 kg benzene

0.56°C



5.12°C/m

Tf



Kf

14-13 Determination of Molecular Weight by Freezing Point Depression or Boiling Point Elevation 567