$1.22 \mathrm{~g}$ of an organic acid is separately dissolved in $100 \mathrm{~g}$ of benzene $\left(\mathrm{K}_{\mathrm{b}}=2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$ and $100 \mathrm{~g}$ of acetone $\left(\mathrm{K}_{\mathrm{b}}=1.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$. 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 $0.17^{\circ} \mathrm{C}$.
The increase in boiling point of solution in benzene
in ${ }^{\circ} \mathrm{C}$ is $\mathrm{x} \times 10^{-2}$. The value of $\mathrm{x}$ is
(Nearest integer)
[Atomic mass: $\mathrm{C}=12.0, \mathrm{H}=1.0, \mathrm{O}=16.0]$
With benzene as solvent
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{K}_{\mathrm{b}} \mathrm{m}$
$\Delta \mathrm{T}_{\mathrm{b}}=\frac{1}{2} \times 2.6 \times \frac{1.22 / \mathrm{M}_{\mathrm{w}}}{100 / 1000}$ ...............(1)
With Acetone as solvent
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{K}_{\mathrm{b}} \mathrm{m}$
$0.17=1 \times 1.7 \times \frac{1.22 / \mathrm{M}_{\mathrm{w}}}{100 / 1000}$ ..................(2)
(1) / ( 2 )
$\frac{\Delta \mathrm{T}_{\mathrm{b}}}{0.17}=\frac{\frac{1}{2} \times 2.6+\frac{1.22 / \mathrm{M}_{\mathrm{w}}}{100 / 1000}}{1 \times 1.7 \times \frac{1.22 / \mathrm{M}_{\mathrm{w}}}{100 / 1000}}$
$\Delta \mathrm{T}_{\mathrm{b}}=\frac{0.26}{2}$
$\Delta \mathrm{T}_{\mathrm{b}}=13 \times 10^{-2}$
$\Rightarrow x=13$
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