If `"0.64 g"` of napthalene (`M = "128.1705 g/mol"`) is dissolved in `"100 g"` a certain solvent (`T_f^"*" = 6^@ "C"`), whose `K_f = 20^@ "C"cdot"kg/mol"`, what is the new freezing point of the solvent?

`(b)`

That's a high `K_f`... If it weren't for the low concentration, `T_f` may have gotten down to `-1^@ "C"`. What would the molality be if the answer were `(a)`?

The change in freezing point is given by:

`DeltaT_f = T_f - T_f^"*" = -iK_fm`,

where:

• `T_f` is the freezing point of the solution.
• `"*"` indicates pure solvent.
• `i` is the van't Hoff factor of the solute. `i >= 1`, and corresponds to the effective number of particles per formula unit of solute.
• `K_f` is the freezing point depression constant of the solvent.
• `m` is the molality of the solution in `"mol solute/kg solvent"`.

`i = 1` for nonelectrolytes, as they essentially do not dissociate in solution. The molality is the next thing we can calculate...

`m = (0.64 cancel"g Napthalene" xx "1 mol"/(128.1705 cancel"g Napthalene"))/(100 cancel"g solvent" xx "1 kg"/(1000 cancel"g solvent")`

`=` `"0.050 mol solute/kg solvent"`

Therefore, the change in freezing point is:

`DeltaT_f = -(1)(20^@ "C"cdot"kg/mol")("0.0499 mol/kg")`

`= -0.9987^@ "C" -> -1^@ "C"`

Thus, the new freezing point is:

`color(blue)(T_f) = DeltaT_f + T_f^"*" = (-0.9987^@ "C") + (6^@ "C")`

`~~ color(blue)(5^@ "C")`