What will be the freezing point of a 20.0% solution of glucose, #C_6H_12O_6# in water?

I got #-2.06^@ "C"#.

I'm assuming #20%# is by mass, not by volume, i.e. it can be written as #"20% w/w"#, or #"20 wt%"#.

Recall the equation for freezing point depression:

#bb(DeltaT_f = T_f - T_f^"*" = -iK_fm_i)#

where:

• #T_f# is the freezing point of the liquid (either solution or solvent), and #""^"*"# indicates/emphasizes that the value is for the pure solvent.
• #i# is the van't Hoff factor, and approximates how many particles per formula unit have dissociated into the solution. For instance, #i = 3# means exactly three ions per solute particle were generated in solution upon dissociation, and that no ion pairing is occurring.
• #K_f = "1.86"^@ "C"cdot"kg"/"mol"# is the freezing point depression constant of water.
• #m_i# is the molality, #"mol solute"/"kg solvent"#, of the solute glucose in the solvent water.
• The negative in front of the far righthand side ensures the molality is positive (as it should be for a physically sensible number).

We can first assume that we have #"100 g"#, or #"0.100 kg"#, of water, so that we know we have #"20 g"# of glucose. That allows us to get a molality of:

#m_i = n_i/w_A = (w_i"/"M_i)/(w_A) = (20 cancel"g glucose"xx "1 mol"/(180.1548 cancel("g glucose")))/("0.100 kg water")#

#= 0.11102/0.100 "mol"/"kg"#

#=# #"1.1102 mol/kg"#

Next, the van't Hoff factor is approximately #i = 1#, since glucose is a non-electrolyte. It dissolves nicely, but does not dissociate well (hardly ionizes at all).

Lastly, we know that #T_f^"*" = 0^@ "C"# for pure water. Therefore, #DeltaT_f = T_f#, and we can solve for #T_f# to get:

#DeltaT_f = -(1)(1.86^@ "C"cdot"kg"/"mol")("1.1102 mol/kg")#

#= color(blue)(-2.06^@ "C" = T_f)#