Assume that you have a solution of an unknown solute in cyclohexane. If the solution has a freezing-point depression of 9.50Celcius, what is the molality of this solution? (The molal freezing-point constant of cyclohexane is 20.2 C/m)

We obtain a molality dependent on the assumed van't Hoff factor of #i ~~ 1#,

#m ~~ 0.470# #"molal"#

We refer to the freezing point depression given by

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

where:

• #T_f# is the freezing point in #""^@ "C"# of the solution, and #"*"# indicates pure solvent.
• #i# is the van't Hoff factor of the solute, the effective number of dissociated particles per formula unit. For nonelectrolytes, this would be #1#, but we have no idea what this solute is...
• #K_f = 20.2^@ "C/m"# is the freezing point depression constant of cyclohexane at its normal freezing point.
• #m# is the molality in #"mol solute/kg solvent"#.

The molality expression is therefore

#color(blue)(m) = -(DeltaT_f)/(iK_f) = -1/i (-9.50^@ "C")/(20.2^@ "C/m")#

#= color(blue)(0.470/i)# #color(blue)("mol solute/kg solvent")#

So we will have to provide a van't Hoff factor to determine the molality here.

Since this is a fairly high molality (high being higher than #"0.01 molal"#), we have to assume the solute is quite soluble, and thus that it is nonpolar and organic, making it a nonelectrolyte.

That means #i ~~ 1#.

Of course, had the change in temperature been low enough that #m < "0.01 molal"#, we would have been in the dark about what kind of solute this could have been, as many things are barely soluble in cyclohexane...