wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solution has 8.35g I2 in 71.0g of toluene. Assuming I2 is nonvolatile. Toluene, Tb=110.630C and Kb=3.400CKg/mol What is the boiling point of solution (in °C) ?

Open in App
Solution

.
ΔTb=iKbm

Where,
i is the van't Hoff factor, i.e. the effective number of dissociated particles per undissociated solute particle.

Kb=3.40C.Kg/mol is the boiling point elevation constant of toluene.

m is the molality, mol solute/kg solvent.

So, let's find the molality of a toluene iodine solution then,

8.35gI2×1mol253.808gI2=0.03290 moles

71.0g toluene = 0.0710kg toluene

Therefore, the molality is:

m=0.03290molesolute0.0710kgsolvent=0.4634mol/kg

i=1.

The change in boiling point is:

ΔTb=(1)(3.40Ckg/mol)(0.4634mol/kg)
=1.58C

As a result, the new boiling point is:
Tb=Tb+ΔTb

=110.63C+1.58C=112.21C

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Liquids in Liquids and Raoult's Law
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon