Question

A solution of 122 gm of benzoic acid in 1000 gm of benzene solvent shows a boiling point elevation of$1.4{}^\circ $. Assuming that the solute gets dimerized to the extent of 80%, then calculate the boiling point of benzene if enthalpy of vaporization of benzene $\Delta {{H}_{vap}}$ = 78 Kcal/mol.

Answer 1

Hint: the boiling point elevation of any solution is directly proportional to the molality of that solution, therefore the elevation in boiling point is$\Delta {{T}_{b}}=i{{K}_{b}}.m$ , where${{K}_{b}}$ is molal elevation constant, i is the Van’t Hoff factor, m is the molality. Another relation is that${{K}_{b}}=\dfrac{R{{T}_{b}}^{2}m}{\Delta {{H}_{vap}}}$ , this includes enthalpy of vaporization.

Complete answer:
We have been given a solution that has 122 gm of benzoic acid as a solute and 1000 gm of benzene as solvent, which results in association to an extent of 80%. We have to calculate the boiling point when elevation in boiling point $\Delta {{T}_{b}}$ is $1.4{}^\circ $. We have been given enthalpy of vaporization of benzene $\Delta {{H}_{vap}}$ as 78 Kcal/mol.
First, we will calculate the molality that is the number of moles of solute upon kilograms of solvent. We have mass of solute so, number of moles = $\dfrac{mass}{molar\,mass}=\dfrac{122}{122}=1\,mole$ (122 g/mol molar mass of benzoic acid). Therefore molality will be:
Molality $m=\dfrac{moles\,of\,solute}{Kg\,of\,solvent}$
$m=\dfrac{1\,mole}{1\,Kg}$, so molality = 1m
Now, as there is dimerization taking place, the number of molecules in product will be 1 and that in reactant 2, so n = 2 and the extent is 80% that means the association is 0.8. So, at association $\alpha =\dfrac{i-1}{\dfrac{1}{n}-1}$ , therefore$0.8=\dfrac{i-1}{\dfrac{1}{2}-1}=0.80\times (-0.5)=i-1$
So, i = 0.6
Now, putting all the respective values in the elevation in boiling point formula$\Delta {{T}_{b}}=i{{K}_{b}}.m$, where we are taking${{K}_{b}}=\dfrac{R{{T}_{b}}^{2}m}{\Delta {{H}_{vap}}}$ we have,
$1.4=\dfrac{0.6\times {{T}_{b}}^{2}\times 78\times 1}{7.8\times {{10}^{3}}\times 78}$ (changing Kcal/mol into Kcal/gm)
${{T}_{b}}^{2}=116.66\times 78$
${{T}_{b}}=95{}^\circ $
Hence, the boiling point ${{T}_{b}}=95{}^\circ $.

Note:
The Van’t Hoff factor is determined by the percentage of the association formula. The kilocalorie per mole is converted to kilo calorie per gram by the factor that 1 Kcal/mol = 1000 Kcal/ gm. Also the molality is taken in kilograms so, 1000 gram is changed into Kg by the factor, 1 Kg = 1000 gm.