Question

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 ?

What is the boiling point (in °C) of a solution of 8.35 g of I2 in 71.0 g of toluene, assuming the I2 is nonvolatile? (For toluene, Tb = 110.63°C and Kb = 3.40°C·kg/mol.)

Answer 1

`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")`