Question

$1.22\, g$ of an organic acid is separately dissolved in $100 \,g$ of benzene $\left( K _{ b }=2.6\, K\, kg\, mol ^{-1}\right)$ and $100\, g$ of acetone $\left( K _{ b }=1.7 \,K \,kg \,mol ^{-1}\right)$. The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by $0.17^{\circ} C$. The increase in boiling point of solution in benzene in ${ }^{\circ} C$ is $x \times 10^{-2}$. The value of $x$ is ______ (Nearest integer) [Atomic mass : $C =12.0, H =1.0, O =16.0$ ]

Answer 1

With benzene as solvent
$\Delta T _{ b }= i K _{ b } m$
$\Delta T _{ b }=\frac{1}{2} \times 2.6 \times \frac{1.22 / M _{ w }}{100 / 1000}....$(1)
With Acetone as solvent
$\Delta T _{ b }= i K _{ b } m $
$0.17=1 \times 1.7 \times \frac{1.22 / M _{ w }}{100 / 1000}.....$(2)
$(1) /(2)$
$\frac{\Delta T _{ b }}{0.17}=\frac{\frac{1}{2} \times 2.6+\frac{1.22 / M _{ w }}{100 / 1000}}{1 \times 1.7 \times \frac{1.22 / M _{ w }}{100 / 1000}}$
$\Delta T _{ b }=\frac{0.26}{2} $
$\Delta T _{ b }=13 \times 10^{-2}$
$\Rightarrow x =13$