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Q.
Given, $H _2 O ( l ) \rightleftharpoons H _2 O ( g )$ at $373 K, \Delta H^{\circ}=8.31\, kcal\,mol ^{-1}$
Thus, boiling point of 0.1 molal sucrose solution is
Solutions
Solution:
$H _2 O$ (l) changes to steam $H _2 O ( g )$ at $373\, K$.
Thus, this represents latent heat of vaporization.
$K _{ b }$ (molal elevation constant) is related to $\Delta H ^{\circ}$ and boiling point by equation,
$K_b=\frac{R T_0^2}{1000 \Delta H^o}$
Here, $\Delta H ^{\circ}$ is in energy unit per gram of solvent.
$\Delta H^{\circ}=8.31\, kcal mol^{-1}$
$=\frac{8.31}{18} kcal\, g^{-1}$
$\therefore K_b=\frac{0.002 \times(373)^2}{1000 \times\left(\frac{8.31}{18}\right)}$
$=\frac{278.25}{461.66}=0.60^{\circ} mol ^{-1} kg$
$\therefore$ Boiling point of solution $=373+0.06^{\circ}=373.06\, K$