PLEASE HELP! If the presence of 2.00g of an unknown non-ionizing substance lowers the freezing temperature of a 10.00g sample of benzene by 6.33 degrees celsius, calculate the molar mass of the substance?

Kf Benzene = 5.12 degrees celsius * kg/mol

1 Answer
Feb 21, 2018

Well, you could simply calculate the first thing that you can get, and go from there. I get about "161.8 g/mol"161.8 g/mol.


The negative change in freezing point is given by

DeltaT_f = -iK_fm < 0

where:

  • i is the van't Hoff factor, i.e. the effective number of dissociated solute particles per undissociated solute particle.
  • K_f = 5.12^@ "C"cdot"kg/mol" (not ""^@ "C"! What happens if you do not have the correct units?) is the freezing point depression constant for benzene. Here we list K_f > 0.
  • m is the molality, i.e. "mols solute"/"kg solvent".

Here we can solve for the molality first.

The change in freezing point is negative, i.e. we have freezing point depression. Furthermore, a non-ionizing solute has i = 1. Therefore:

m = (DeltaT_f)/(-iK_f)

= (-6.33^@ "C")/(-(1)(5.12^@ "C"cdot"kg/mol")

= "1.236 mol solute/kg solvent"

Therefore, if we have "2.00 g solute" and "10.00 g benzene"... well, benzene is the solvent. It has to be, because K_f can only be given for the solvent.

The molality, an intensive property, is fixed for whatever choice of solute mass and solvent mass we pick. So...

"1.236 mol solute"/"kg solvent" = ("mol solute"/"??? g solute" xx "2.00 g solute")/("0.01000 kg solvent")

Therefore:

1/["1.236 mol solute"/cancel"kg solvent" xx (0.01000 cancel"kg solvent")/"2.00 g solute"] = "??? g"/"mol solute"

And we have color(blue)"161.8 g/mol" for the molar mass.