What mass of sucrose was dissolved in #"100 g"# of water to lower its freezing point by #1^@ "C"#? #K_f = -1.86^@ "C/m"#.

I got #"18.40 g sugar"#, but I can only report #"20 g"#.

Using your convention that #K_f < 0#, you seem to be using:

#DeltaT_f = iK_fm#

where:

• #i# is the van't Hoff factor, i.e. the effective number of particles that dissociate for every solute particle.
• #K_f = -1.86^@ "C"cdot"kg/mol"# is the freezing point depression constant for water.
• #m# is the molality of the solution in #"mols solute/kg solvent"#.

Sugar is a nonelectrolyte, so every particle of sugar stays pretty much as sugar without breaking apart, and simply dissolves without dissociating.

Therefore, #i = 1#, and if the decrease in freezing point is #-1^@ "C"#, then the set up is:

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

So, the molality is:

#=> m = (-1^@ "C")/((1)(-1.86^@ "C"cdot"kg/mol"))#

#=# #"0.5376 mols solute/kg solvent"#

At this point, you know that you have #"100 g"# of water, the solvent... so you have #"0.100 kg"# of water. Therefore, you have this many mols of solute:

#"0.5376 mols solute"/cancel"1 kg solvent" xx 0.100 cancel"kg solvent"#

#=# #"0.05376 mols solute"#

Your solute was sugar, which is generally regarded as sucrose, #"C"_12"H"_22"O"_11#, with molar mass

#12 cdot 12.011 + 22 cdot 1.0079 + 11 cdot 15.999 "g/mol"#

#=# #"342.2948 g sucrose/mol"#.

Therefore, this mass of sucrose was needed to decrease the freezing point of the sugar solution, relative to pure water, by #1^@ "C"#:

#0.05376 cancel"mols sucrose" xx "342.2948 g sucrose"/cancel"mol sucrose"#

#=# #color(blue)ul("18.40 g sucrose/sugar")#

But based on the numbers you gave, I can only state #"20 g"# for one significant figure...