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"#.)

Question

2055 views around the world

You can reuse this answer
Creative Commons License

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=??#