# Question b576d

Feb 4, 2017

The depression of freezing point of solvent $\Delta {T}_{f}$ is related with the molality $m$ of the solute as follows

$\textcolor{b l u e}{\Delta {T}_{f} = - i \cdot {K}_{f} \cdot m \ldots \ldots \ldots . \left[1\right]}$

where

${K}_{f} \to \text{cryoscopic constant "=1.853Kkg"/mol for water}$

$m \to \text{molality of the soute} = 0.100$

$i \to \text{ van't hoff factor} = 1 + \alpha$

alpha->"degree of dissociation"=1.32%=0.0132#

Replacing i by $1 + \alpha$ the equation [1] becomes

$\textcolor{g r e e n}{\Delta {T}_{f} = - \left(1 + \alpha\right) \cdot {K}_{f} \cdot m}$

$\textcolor{g r e e n}{\implies \Delta {T}_{f} = - \left(1 + 1.0132\right) \cdot 1.853 \cdot 0.1 K \approx - 0.188 K}$

So expected freezing point will be

$\textcolor{red}{\text{Expected freezing point} = \left(273 - 0.188\right) K = 272.812 K}$