What is the change in freezing point when #"54 g"# of glucose (use #"FW" = "180 g/mol"#) is dissolved in #"0.250 kg"# of water? #K_f = 1.86^@ "C/m"# for water.

#-2.23^@ "C"#

Well, the fact is, we (usually) don't care what the solute is. All we needed was how many #"mols"# of it we have (which we can get regardless of what its name is), whether it's a strong electrolyte or not, and how many #"kg"# of water there are.

The decrease in freezing point is given by:

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

where:

• #T_f# is the freezing point of the solution.
• #T_f^"*"# is the freezing point of the pure solvent.
• #i# is the van't Hoff factor of the solute, i.e. the number of particles per formula unit that have gone into solution upon any dissociation.
• #K_f = 1.86^@ "C/m"#, or #""^@ "C"cdot"kg/mol"# is the freezing point depression constant of water.
• #m# is the molality of the solution in #"mol solute/kg solvent"#.

Given the mass of the glucose, we have:

#54 cancel"g solute" xx "1 mol"/(180 cancel"g glucose")#

#=# #"0.30 mols glucose"#

And we know that #"250 g"# is #"0.250 kg"# of water. Water is, of course, the solvent. So:

#m = "0.30 mols glucose"/"0.250 kg water"#

#=# #"1.20 mol/kg"#

Therefore, noting that glucose is a NONelectrolyte (i.e. its van't Hoff factor is #1#), the change in freezing point is:

#DeltaT_f = -(1)(1.86^@ "C"cdot"kg/mol")("1.20 mol/kg")#

#= -2.23^@ "C"#

Therefore, if we wished, we could easily find the new freezing point.

#T_f - T_f^"*" = DeltaT_f#

#=> color(blue)(T_f) = DeltaT_f + cancel(T_f^"*")^(0^@ "C")#

#= color(blue)(-2.23^@ "C")#

Naturally, it's the same value as the change in freezing point, since the freezing point of water is #0^@ "C"# at #"1 atm"#.