# 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.

##### 1 Answer

#-2.23^@ "C"#

Well, the fact is, we (usually) don't care what the solute is. All we needed was how many

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 thefreezing point depression constantof water.#m# is themolalityof 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

#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 **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