How many grams of sucrose should be dissolved in 100 g water in order to produce a solution with a 105°C difference between freezing point and boiling point temperatures? (M.Wt sucrose = 342, Kb=0.51, Kf=1.86)

I got #"72.2 g sucrose"#.

What percent of the #5^@ "C"# expansion of the transition temperature range is contributed by the boiling point elevation? (You should get about #22%#.)

Well, boiling point elevation and freezing point depression are given by...

#DeltaT_b = T_b - T_b^"*" = iK_bm#

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

where

• #T_(tr)# is the phase transition temperature of the solvent in the context of the solution, and #"*"# indicates pure solvent.
• #i# is the van't Hoff factor, i.e. the effective number of dissociated particles per solute particle placed into solvent.
• #K_b = 0.512^@ "C"cdot"kg/mol"# is the boiling point elevation constant.
• #K_f = 1.86^@ "C"cdot"kg/mol"# is the freezing point depression constant.
• #m = "mols solute"/"kg solvent"# is the molality of the solution.

Any solute added to a solvent will expand its boiling point to freezing point range.

If you want the gap to be #105^@ "C"# between the new #T_f# and #T_b#, then...

#T_b - T_f = 105^@ "C"#

#= DeltaT_b + T_b^"*" - (DeltaT_f + T_f^"*")#

#= DeltaT_b - DeltaT_f + T_b^"*" - T_f^"*"#

We know that the boiling point and freezing point of water are #100^@ "C"# and #0^@ "C"# respectively, so...

#105^@ "C" = DeltaT_b - DeltaT_f + 100^@ "C"#

or

#5^@ "C" = DeltaT_b - DeltaT_f#

#= iK_bm + iK_fm#

#= im(K_b + K_f)#

Therefore, as sucrose is a nonelectrolyte, #i = 1# and the required molality of the solution is:

#m = (5^@ "C")/(i(K_b + K_f))#

#= (5^@ "C")/(1 cdot (0.512^@ "C"cdot"kg/mol" + 1.86^@ "C"cdot"kg/mol")#

#=# #"2.108 mols solute/kg solvent"#

And since we have #"100 g solvent"#...

#"100 g water" -> "0.100 kg water"#

Thus, the mols of sucrose in the water is:

#"2.108 mols solute"/cancel"kg solvent" xx 0.100 cancel"kg solvent"#

#=# #"0.2108 mols sucrose"#

And this mass is

#w_"sucrose" = 0.2108 cancel("mols C"_12"H"_22"O"_11) xx "342.295 g sucrose"/cancel"1 mol sucrose"#

#=# #color(blue)("72.2 g sucrose")#