Question

Given, $H_{2}O(l)\rightleftharpoons H_{2}O(g)$ at $373K,\Delta H^{\circ }=8.31kcalmo{l}^{-1}$ Thus, boiling point of 0.1 molal sucrose solution is

• A. 373.52 K
• B. 373.052 K
• C. 373.06 K
• D. 374.52 K

Answer 1

Answer: (C)

$H_{2}O$ (l) changes to steam $H_{2}O(g)$ at $373K$.
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.
$\Delta H^{\circ }=8.31kcalmo{l}^{-1}$
$=\frac{8.31}{18}kcal{g}^{-1}$
$\therefore K_b=\frac{0.002\times (373{)}^2}{1000\times \left(\frac{8.31}{18}\right)}$
$=\frac{278.25}{461.66}=0.60^{\circ }mo{l}^{-1}kg$
$\therefore$ Boiling point of solution $=373+0.06^{\circ }=373.06K$