# When "18 g" of ethylene glycol (C_2H_6O_2) is dissolved in "150 g" of pure water, what is the freezing point of the solution? (The freezing point depression constant for water is 1.86^@ "C"cdot"kg/mol".)

Aug 12, 2017

Approx. $- 3.6$ ""^@C

#### Explanation:

Ethylene glycol is a molecular species, and we need to calculate the molality of its water solution.....

$\text{Molality"="Moles of solute"/"Kilograms of solvent} = \frac{\frac{18 \cdot g}{62.07 \cdot g \cdot m o {l}^{-} 1}}{\frac{150 \cdot g}{1000 \cdot g \cdot k {g}^{-} 1}} = 1.93 \cdot m o l \cdot k {g}^{-} 1$

DeltaT_f=k_fxx1.93*mol*kg^-1=(1.86*""^@C*kg)/(mol)xx1.93*(mol)/(kg^-1)

$\Delta {T}_{f} = 3.59$ ""^@C.

And this is freezing point depression, and thus the observed freezing point of the solution is $\left(0 - 3.6\right)$ ""^@C=??