http://www.ck12.org Chapter 16. Solutions
6 .44 mol C 2 H 6 O 2
0 .500 kg H 2 O
= 12. 9 mC 2 H 6 O 2
∆Tf= Kf×m= -1.86°C/m×12.9m= -24.0°C
Tf= -24.0°CThe normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing
point of the solution is−24.0°C.
Step 3: Think about your result.
The freezing point of the water decreases by a large amount, protecting the radiator from damage due to the
expansion of water when it freezes. There are three significant figures in the result.
Practice Problem- Calculate the freezing point of a solution prepared by dissolving 27.56 g of glucose (C 6 H 12 O 6 ) into 125 g of
water.
Colligative properties have practical applications, such as the salting of roads in cold-weather climates (Figure
16.12). By applying salt to an icy road, the melting point of the ice is decreased, and the ice will melt more quickly,
making driving safer. Sodium chloride (NaCl) and, either calcium chloride (CaCl 2 ), or magnesium chloride (MgCl 2 ),
are used most frequently, either alone or in a mixture. Sodium chloride is the least expensive option, but it is less
effective because it only dissociates into two ions instead of three.
FIGURE 16.12
Workers are pouring salt onto an icy road
surface in order to lower the melting point
of the ice.Boiling Point Elevation
The figure below (Figure16.13) shows, again, the phase diagram of a solution and the effect that the lowered vapor
pressure has on the boiling point of the solution compared to the solvent. In this case, the solution has a higher
boiling point than the pure solvent. Since the vapor pressure of the solution is lower, more heat must be supplied to
the solution to bring its vapor pressure up to the pressure of the external atmosphere. Theboiling point elevation
is the difference in temperature between the boiling point of the pure solvent and that of the solution. On the graph,
the boiling point elevation is represented by∆Tb.