Question

A solution containing $0.122kg$ of benzoic acid in $1kg$ of benzene boils (boiling point $353K$) at $354.5K$.
Determine the apparent molar mass of benzoic acid (which dimerizes) in the solution and degree of dimerization.
Given: ${{\Delta }_{vap}}{{H}_{1m}}(benzene)=394.57\,J\,{{g}^{-1}}$
(A)- $0.214\,kg\,mo{{l}^{-1}},0.86$
(B)- $0.257\,kg\,mo{{l}^{-1}},0.932$
(C)- $0.428\,kg\,mo{{l}^{-1}},0.956$
(D)- $0.454\,kg\,mo{{l}^{-1}},0.986$

Answer 1

The colligative property is dependent on the number of solute particles and the extent of association or dissociation of the solute particles as they dissolve with the solvent particles. The boiling point increases, on adding the solute.

Complete step by step answer:
In the given system, an elevation in boiling point can be seen as the boiling point of the solution is greater than the boiling point of the solvent (benzene). That is, an elevation of $(354.5-353=)1.5K$ is seen in the solution compared to the solvent.
Then applying the colligative property of elevation in boiling point, we have the boiling point elevation as:
$\Delta {{T}_{b}}=im{{K}_{b}}$ ---------- (a)
where, $i=$ Van’t Hoff factor
$m=$ molality of the solution
${{K}_{b}}=$Ebullioscopic constant which is dependent on the solvent
${{K}_{b}}=R{{T}_{b}}^{2}M/\Delta {{H}_{vap}}$ ---------- (b)
where $R$ is the gas constant $(8.314\,Jmo{{l}^{-1}}{{K}^{-1}})$ , $M$ is the molar mass of the solvent, ${{T}_{b}}$ is
the temperature of solvent in Kelvin and $\Delta {{H}_{vap}}$ is the heat of vaporisation per mole of solvent.
Then, we are given:${{\Delta }_{vap}}{{H}_{1m}}(benzene)=394.57\,J\,{{g}^{-1}}=394.57\times 78\,Jmo{{l}^{-1}}$ and $M=78g/mol=0.078kg/mol$
So, substituting these values in equation (b), we get,
${{K}_{b}}=\dfrac{8.314\times {{\left( 353 \right)}^{2}}\times 0.078}{394.5\times 78}$
$=2.626K\,kg{{\,}^{{}}}mo{{l}^{-1}}$
Then, substituting the value of ${{K}_{b}}$ in equation (a), we get,
Molality of the solution$=m=\dfrac{{{T}_{b}}}{{{K}_{b}}}=\dfrac{1.5}{2.626}=0.57\,mol/kg$
Then, molar mass of solute (benzoic acid) will be,
Molality $=0.57=\dfrac{\text{moles}\,\text{of}\,\text{solute}}{\text{mass}\,\text{of solvent in kg}}=\dfrac{0.122}{\text{molar mass}\times 1}$
Then, the apparent molar mass of benzoic acid $=\dfrac{0.122}{0.57}=0.214kg/mol$
In the given solution of benzoic acid in benzene, the benzoic acid dimerises. Let the degree of dimerization be $\alpha$. Then,
$\underset{(1-\alpha )}{\mathop{{{C}_{6}}{{H}_{5}}-COOH}}\,\rightleftarrows \underset{(\alpha /2)}{\mathop{{{({{C}_{6}}{{H}_{5}}-COOH)}_{2}}}}\,$
Then, total amount of solute $=(1-\alpha +\alpha /2)=(1-\alpha /2)\,mol$
So, the apparent molar mass of solute $=0.214\,kg/mol=\dfrac{\text{mass of solute added}}{\text{total mass of solute}}$
$0.214=\dfrac{0.122}{(1-\alpha /2)}$

Then, we get the degree of dimerization, $\alpha =0.86$.

Therefore, the apparent molar mass of benzoic acid (which dimerizes) in the solution and degree of dimerization is option (A)- $0.214\,kg\,mo{{l}^{-1}},0.86$.

Note: The solution has a higher boiling point than the pure solvent (benzene) because in the presence of the non-volatile solute (benzoic acid), which does not contribute to the vapor pressure of the solution. So, the higher temperature makes solvent to exert pressure on the solution vapour pressure.