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 $1.86^@ "C"cdot"kg/mol"$ .)

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Answer 1 · 2017-08-12T05:26:01

Approx. $-3.6$ $""^@C$

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

$"Molality"="Moles of solute"/"Kilograms of solvent"=((18*g)/(62.07*g*mol^-1))/((150*g)/(1000*g*kg^-1))=1.93*mol*kg^-1$

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

$DeltaT_f=3.59$ $""^@C$ .

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

When "18 g""18 g" of ethylene glycol (C_2H_6O_2C_2H_6O_2) is dissolved in "150 g""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"1.86^@ "C"cdot"kg/mol".)