Cracking the SAT Chemistry Subject Test

BOILING POINT ELEVATION AND FREEZING

POINT DEPRESSION

When a solute is dissolved in a liquid solvent, the solvent’s boiling point is
raised, its freezing point is lowered, and its vapor pressure is lowered. By how
much? The change in boiling point (∆Tb) or freezing point (∆Tf) is always equal

to a constant (k) times the number of moles of dissolved particles of solute per
kilogram of solvent.

∆T  =   kmi

The value k is different for different solvents; however, for all liquid solvents,
the extent of boiling point elevation or freezing point depression is directly
proportional to the molality of the solute. It is also directly proportional to the
number of dissolved particles, which we denote by i. For example, when we
dissolve NaCl in water, it dissociates into Na+ ions and Cl− ions. For every mole
of NaCl we dissolve, we get 1 mole of Na+ and 1 mole of Cl−, for a total of 2
moles of dissolved particles. For NaCl, then, i = 2. If we dissolve 1 mole of
sucrose (table sugar), however, the sugar molecules dissolve but don’t dissociate,
so we get only 1 mole of dissolved particles, and i = 1.

Now, suppose we take the same amount of water and add 1 mole of KCl. KCl is
an ionic substance; it dissociates in solution. Each unit of KCl produces two
dissolved particles: 1 K+ ion and 1 Cl− ion. The boiling point elevation and
freezing point depression will be twice as great as they would be in the case of
the dissolution of sucrose.

Let’s   Look    at  That    Equation    Again

ΔT  =   kmi

k   =   constant    that    depends on  solvent