What mass of ethylene glycol( non -volatile, non ionizing, 62.1g/mole) is neededto prepare 10.0L water(#rho = "1 g/mL"#) with a freezing point of ( -23 degree celcious) ?

1 Answer
Truong-Son N.
Mar 28, 2018

I get #"7.68 kg"#.

Since the freezing point has to decrease, this is referring to freezing point depression:

#DeltaT_f = -iK_fm#

where:

  • #i = 1# is the van't Hoff factor for non-ionizing solutes.
  • #K_f = 1.86^@ "C/m"# is the freezing point depression constant of water.
  • #m# is the molality of the solution in #"mol solute/kg solvent"#.

So you can solve for the molality first.

#m = -(DeltaT_f)/(iK_f)#

#= -(-23^@ "C")/(1 cdot 1.86^@ "C/m") = "12.37 mol EG/kg water"#

Thus, if you want #"10.0 L"# water, and water is assumed to have a density of #"1 g/mL"#, it is #"10.0 kg"#. Therefore, you have #"123.7 mols EG"#, or

#123.7 cancel"mols EG" xx "62.1 g"/cancel"mol" = 7.68 xx 10^3 "g"#

#=# #color(blue)("7.68 kg ethylene glycol")#

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