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
Truong-Son N.
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"#.

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.

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