Question

A solution containing $0.122kg$ of benzoic acid in $1kg$ of benzene (b.pt.353 K) boils at $354.5K$.
Determine the apparent molar mass of benzoic acid (which dimerizes) in the solution and the degree of dimerization.
Given: ${\Delta }_{vap}{H}_{1m}\left(benzene\right)=394.57J{g}^{-1}$.

• A. $0.214kgmo{l}^{-1},0.86$
• B. $0.257kgmo{l}^{-1},0.932$
• C. $0.428kgmo{l}^{-1},0.956$
• D. $0.454kgmo{l}^{-1},0.986$

Answer 1

The correct option is A $0.214kgmo{l}^{-1},0.86$

In benzene, $k_b=\frac{M_{1}RT{T}_b^2}{{\Delta }_{vap}H}=\frac{0.078\times 8.314\times (353{)}^2}{394.57\times 78}=2.626Kkg/mol$
The molality of the solution is $m=\frac{\Delta T_b}{K_b}=\frac{354.5-353}{2.626}=0.57mol/kg$
$M_2=\frac{m_2}{m{m}_1}=\frac{0.122}{0.57\times 1}=0.214$
Let a be the degree of dimerization of benzoic aicd.
The solution contains 1 mole of benzoic acid.
$\underset{(1-a)mol}{PhCOOH}\rightleftharpoons \underset{\left(a/2\right)mol}{(PhCOOH{)}_2}$
Total amounts of solutes $=(1-a+a/2)mol=(1-a/2)mol$
The number average molar mass of solutes in solution is
$\frac{m}{n_1+n_2}=\frac{0.122kg}{(1-a/2)mol}$
Equate this to 0.214 kg/mol
$\frac{0.122}{1-a/2}=0.214$
$a=2(1-0.122/0.214)=0.86$