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

##### 1 Answer

I got

I'm assuming

Recall the equation for freezing point depression:

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

#T_f# is thefreezing pointof the liquid (either solution or solvent), and#""^"*"# indicates/emphasizes that the value is for the pure solvent.#i# is thevan't Hoff factor, and approximates how manyparticles per formula unithave 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 thefreezing point depression constantof water.#m_i# is themolality,#"mol solute"/"kg solvent"# , of thesoluteglucose in thesolventwater.- 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

#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 **does not** dissociate well (hardly ionizes at all).

Lastly, we know that

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

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