Boiling Point Help! Solution has 8.35g l2 in 71.0g of toluene. Assuming I2 is nonvolatile. Toluene, Tb = 110.63celsius and Kb = 3.40celsius*kg/mol ?

#T_b = 112.21^@ "C"#

What was the change in boiling point?

Well, the change in boiling point is given by...

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

where:

• #i# is the van't Hoff factor, i.e. the effective number of dissociated particles per undissociated solute particle.
• #K_b = 3.40^@ "C"cdot"kg/mol"# is the boiling point elevation constant of toluene.
• #m# is the molality, #"mol solute/kg solvent"#.

Where's your solvent here? It must be toluene, right? You were given #K_b# for toluene. So, let's find the molality of a toluenic iodine solution then.

#8.35 cancel("g I"_2) xx "1 mol"/(253.808 cancel("g I"_2)) = "0.03290 mols"#

#"71.0 g toluene"# #=# #"0.0710 kg toluene"#

Therefore, the molality is

#m = "0.03290 mols solute"/"0.0710 kg solvent" = "0.4634 mol/kg"#

Clearly, iodine doesn't break apart in toluene to form #"I"^(+)# and #"I"^(-)# ions. That's just weird. Toluene is a hydrocarbon, with dispersion forces... Therefore, #i = 1#.

The change in boiling point is...

#DeltaT_b = (1)(3.40^@ "C"cdotcancel"kg/mol")(0.4634 cancel"mol/kg")#

#= 1.58^@ "C"#

As a result, the new boiling point is:

#color(blue)(T_b) = T_b^"*" + DeltaT_b#

#= 110.63^@ "C" + 1.58^@ "C" = color(blue)(112.21^@ "C")#