# For an arbitrary nonelectrolyte, calculate the freezing point when #"0.0192 mols"# of it is dissolved in #"5.00 g"# of water? #K_f = 1.86^@ "C/m"# for water.

##### 1 Answer

Jan 26, 2017

The idea is that dissolving anything into a solvent will decrease its freezing point. It is expected that you already know *final, lowered freezing point*.

Recall:

#DeltaT_f = T_f - T_f^"*" = -iK_fm# where:

#i = 1# is thevan't Hoff factorfor a non-electrolyte. It simply says that only one nonelectrolyte particle exists for every nonelectrolyte particle that is placed into solution.#K_f = 1.86^@ "C/m"# is thefreezing point depression constantforwater.#m# is themolalityof thesolution, i.e.#"mol solute"/"kg solvent"# .#T_f# is thefreezing pointof the.solution#T_f^"*"# is thefreezing pointof the.pure solvent

First, calculate what

#DeltaT_f = -(1)(1.86^@ "C/m")(("0.0192 mols solute")/(5.00xx10^(-3) "kg H"_2"O"))#

#= -7.14^@ "C"#

Therefore, the final freezing point is:

#color(blue)(T_f) = DeltaT_f + T_f^"*" = -7.14^@ "C" + 0^@ "C" = color(blue)(-7.14^@ "C")#