Question #3d506

1 Answer
Dec 29, 2017

Answer:

#0.146#

Explanation:

The first thing that you need to do here is to figure out the molality of this solution. To do that, use the equation

#color(blue)(ul(color(black)(DeltaT_"f" = i * K_f * b)))#

Here

  • #T_"f"# is the freezing point depression
  • #i# is the van't Hoff factor
  • #b# is the molality of the solution
  • #K_f# is the cryoscopic constant of water, equal to# 1.86^@"C kg mol"^(-1)#

Now, the freezing point depression is defined as the difference between the normal freezing point of the pure solvent and the freezing point of the solution.

As you know, pure water has a normal freezing point of #0^@"C"#. Since your solution is said to have a freezing point of #-17.6^@"C"#, you can say that the freezing point depression is

#DeltaT_"f" = 0^@"C" - (-17.6^@"C")#

#DeltaT_"f" = 17.6^@"C"#

You're looking for the molality of the solution, so rearrange the equation to solve for #b#

#DeltaT_"f" = i * K_f * b implies b = (DeltaT_"f")/(i * K_f)#

Since ethylene glycol is a non-electrolyte, i.e. a compound that does not dissociate to form ions in aqueous solution, its van't Hoff factor will be equal to #1#.

This means that you have

#b = (17.6 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "9.46 mol kg"^(-1)#

This means that this solution contains #9.46# moles of ethylene glycol for every #"1 kg"# of water, the solvent.

To find the mole fraction of ethylene glycol in this solution, pick a sample that contains exactly #"1 kg" = 10^3color(white)(.)"g"# of water. Use the molar mass of water to determine how many moles of water are present in this sample.

#10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"#

The total number of moles present in this sample will be

#"9.46 moles + 55.51 moles = 64.97 moles"#

This means that the mole fraction of ethylene glycol, which is defined as the number of moles of ethylene glycol divided by the total number of moles present in the sample, is equal to

#color(darkgreen)(ul(color(black)(chi_ ( ("CH"_ 2"OH")_ 2)))) = (9.46 color(red)(cancel(color(black)("moles"))))/(64.97color(red)(cancel(color(black)("moles")))) = color(darkgreen)(ul(color(black)(0.146)))#

The answer is rounded to three sig figs, the number of sig figs you have for the freezing point of the solution.

Notice that you don't need to know the percent concentration of the solution because the only relevant piece of information here is the freezing point of the solution.