**Practice Problem 7**

Determine the molecular weight of acetic acid if a solution
that contains 30.0 grams of acetic acid per kilogram of water
freezes at -0.93^{o}C. Do these results agree with the
assumption that acetic acid has the formula CH_{3}CO_{2}H?

**Solution**

The freezing point depression for this solution is equal to
the difference between the freezing point of the solution (-0.93^{o}C)
and the freezing point of pure water (0^{o}C).

*T*_{FP} = -0.93^{o}C -
0.0^{o}C = -0.93^{o}C

We now turn to the equation that defines the relationship between the freezing point depression and the molality of the solution.

*T*_{FP} = - *k*_{f}*m*

Since we know the change in the freezing point, and we can look up the freezing point depression constant in a table, we have enough information to calculate the molality of the solution.

m |
= | T_{FP} |
= 0.93^{o}C
= |
0.50 m |

k_{f} |
1.853^{o}m |

At this point, we might return to the statement of the problem, to see if we are making any progress toward an answer. According to this calculation, there are 0.50 moles of acetic acid per kilogram of water in this solution. The problem stated that there were 30.0 grams of acetic acid per kilogram of water in the solution. Since we simultaneously know the number of grams and the number of moles of acetic acid in this sample, we can calculate the molecular weight of acetic acid.

30.0 g |
= | 60 g/mol |

0.50 mol |

The results of this experiment are in good agreement with the
molecular weight (60.05 g/mol) expected if the formula for acetic
acid is CH_{3}CO_{2}H.