Question

Calculate the boiling point of a solution containing $0.61$ g of benzoic acid in $50$ g of carbon disulphide assuming $84$ percent dimerization of the acid. The boiling point and ${{\text{K}}_{\text{b}}}$ of ${\text{C}}{{\text{S}}_{\text{2}}}$ are ${\text{46}}{\text{.2}}{\,^{\text{o}}}{\text{C}}\,$ and ${\text{2}}{\text{.3}}\,{\text{K}}\,{\text{Kg}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}$ respectively.
A. ${\text{46}}{\text{.33}}{\,^{\text{o}}}{\text{C}}\,$
B. ${\text{90}}{\text{.4}}{\,^{\text{o}}}{\text{C}}\,$
C. ${\text{12}}{\text{.3}}{\,^{\text{o}}}{\text{C}}\,$
D. ${\text{62}}{\text{.21}}{\,^{\text{o}}}{\text{C}}\,$

Answer 1

The elevation in boiling point is the product of the van't Hoff factor, boiling point elevation constant, and molality. We will determine the degree of association first. Then we will determine the molality by using the molality formula. Then by using the elevation in boiling point formula we will determine the elevation in boiling point. Then by adding it to the boiling point of pure solvent we will determine the boiling point of the solution.

Complete step by step solution:
The formula of boiling point elevation is as follows:
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = i}}\,{{\text{K}}_{\text{b}}}{\text{.}}\,{\text{m}}$
Where,
${\delta }{{\text{T}}_{\text{b}}}\,$ is the elevation in boiling point.
${{\text{K}}_{\text{b}}}$ is the boiling point elevation constant.
${\text{i}}\,$ is the van't Hoff factor.
The formula to determine the van't Hoff factor for the association is as follows:
${\text{i}}\, = \,1 - \left( {1 - \dfrac{{\text{1}}}{{\text{n}}}} \right){\alpha }$
Where,
${\alpha }$ is the degree of association
${\text{n}}$ is the number of atoms undergoing polymerization.

On dissolving benzoic acid in carbon disulphide, the benzoic acid undergoes association which is shown as follows:
${\text{2}}\,{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\,\,\mathop \rightleftharpoons \limits^{{\text{C}}{{\text{S}}_2}} \,\,{({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH)}}_2}$
Two molecules are undergoing dimerization and the degree of association is $84$ %.
So, the value of n is $2$ .
On substituting $2$ for n and $84$ % for ${\alpha }$ ,
$\Rightarrow {\text{i}}\, = \,1 - \left( {1 - \dfrac{{\text{1}}}{2}} \right)\dfrac{{84}}{{100}}$
$\Rightarrow {\text{i}}\, = \,1 - \left( {\dfrac{{\text{1}}}{2}} \right)0.84$
$\Rightarrow {\text{i}}\, = \,1 - 0.42$
$\Rightarrow 0.58$

Now we will determine the molarity as follows:
${\text{molality}}\,{\text{ = }}\,\dfrac{{{\text{mole}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Kg}}\,{\text{of}}\,{\text{solvent}}}}$
To calculate the molality we have to determine the mole of solute (benzoic acid) first.
The mole formula is as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
Molar mass of benzoic acid is $122$ g/mol.
On substituting $0.61$ g for mass and $122$ g/mol for molar mass,
$\Rightarrow {\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.61}}}}{{{\text{122}}}}$
$\Rightarrow {\text{mole}}\,{\text{ = }}\,{\text{0}}{\text{.005}}$
So, moles of benzoic acid are ${\text{0}}{\text{.005}}$ .

We will convert the kg of solvent (carbon disulphide) from gram to kilogram as follows:
${\text{1000}}\,{\text{g}}\,{\text{ = }}\,{\text{1}}\,{\text{Kg}}$
${\text{50}}\,\,{\text{g}}\,{\text{ = }}\,{\text{0}}{\text{.05}}\,{\text{Kg}}$
So, the mass of solvent is ${\text{0}}{\text{.05}}$ kg.
On substituting ${\text{0}}{\text{.05}}$ kg for kg of solvent and ${\text{0}}{\text{.005}}$ for moles of solute,
$\Rightarrow {\text{molality}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.005}}\,{\text{mol}}\,}}{{{\text{0}}{\text{.05}}\,{\text{Kg}}}}$
$\Rightarrow {\text{molality}}\,{\text{ = }}\,{\text{0}}{\text{.1}}\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}$

Now we will use elevation in the boiling point formula to determine the boiling point elevation.
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = i}}\,{{\text{K}}_{\text{b}}}{\text{.}}\,{\text{m}}$
On substituting, $0.58$ for ${\text{i}}\,$ , $\Rightarrow{\text{2}}{\text{.3}}\,{\text{K}}\,{\text{Kg}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}$ for ${{\text{K}}_{\text{b}}}$ and ${\text{0}}{\text{.1}}\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}$ for. ${\text{m}}$
$\Rightarrow {\delta }{{\text{T}}_{\text{b}}}\,{\text{ = 0}}{\text{.58}}\,\, \times {\text{2}}{\text{.3}}\,{\text{K}}\,{\text{Kg}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{.}}\,{\text{0}}{\text{.1}}\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}$
$\Rightarrow {\delta }{{\text{T}}_{\text{b}}}\,{\text{ = }}0.1334\,{\text{K}}$
$\Rightarrow {\delta }{{\text{T}}_{{\text{b}}\,}}$ is the temperature difference.
The difference will remain in the kelvin unit of degree Celsius so, we can write it as $0.1334{\,^{\text{o}}}{\text{C}}$ .

The relation between the boiling point of the solution and the boiling point of the pure solvent is as follows:
${\delta }{{\text{T}}_{{\text{b}}\,}}\,{\text{ = }}\,{{\text{T}}_{\text{b}}}\, - {\text{T}}_{\text{b}}^{\text{o}}$
Where,
${{\text {T}}_{\text{b}}}$ is the boiling point of the solution
${\text {T}}_{\text{b}}^{\text{o}}$ is the boiling point of the pure solvent
On substituting $0.1334{\,^{\text{o}}}{\text{C}}$ for ${\delta}{{\text{T}}_{{\text{b}}\,}}$ and ${\text{46}}{\text{.2}}{\,^{\text{o}}}{\text{C}}\,$ for ${\text{T}}_{\text{b}}^{\text{o}}$ ,
$\Rightarrow {\text{0}}{\text{.1334}}{\,^{\text{o}}}{\text{C}}\,{\text{ = }}\,{{\text{T}}_{\text{b}}}\, - {\text{46}}{\text{.2}}{\,^{\text{o}}}{\text{C}}\,$
$\Rightarrow {{\text{T}}_{\text{b}}}\,{\text{ = }}\,\,{\text{0}}{\text{.1334}}{\,^{\text{o}}}{\text{C}}\, + {\text{46}}{\text{.2}}{\,^{\text{o}}}{\text{C}}\,$
$\Rightarrow {{\text{T}}_{\text{b}}}\,{\text{ = }}\,\,46.33{\,^{\text{o}}}{\text{C}}\,$
So, the boiling point of a solution is ${\text{46}}{\text{.33}}{\,^{\text{o}}}{\text{C}}\,$ .

Therefore, option (A) is correct.

Note: The van't Hoff factor represents the degree of dissociation or association of the molecule in the solution. The number of ions, or molecules at equilibrium represents the value of van't Hoff factor. According to the boiling point elevation concept, on adding the non-volatile solute in a solvent the vapour pressure of the solution decreases so the boiling point of the solution increases.So, the boiling point of the solution will be the sum of boiling point of the pure solvent and boiling point elevation.