Question #e16a6

1 Answer
Sep 3, 2017

Answer:

Here's what I got.

Explanation:

The key here is the difference between the ebullioscopic constant, #K_b#, and the cryoscopic constant, #K_f#, of water. the solvent of the solution.

You know that you have--see here for the source

  • #K_b = 0.512^@"C kg mol"^(-1)#
  • #K_f = 1.86^@"C kg mol"^(-1)#

Now, the boiling-point elevation is calculated using

#DeltaT_"b" = i * K_b * b#

Here

  • #i# is the van't Hoff factor
  • #b# is the molality of the solution

Similarly, the freezing-point depression is calculated using

#DeltaT_"f" = i * K_f * b#

Since you're dealing with a single solution here, i.e. the same molality and van't Hoff factor, you can divide the two equations to get

#(DeltaT_"b")/(DeltaT_"f") = (color(red)(cancel(color(black)(i))) * K_b * color(red)(cancel(color(black)(b))))/(color(red)(cancel(color(black)(i))) * K_f * color(red)(cancel(color(black)(b))))#

This means that you have

#DeltaT_"f" = K_"f"/K_"b" * DeltaT_"b"#

Now, water has a normal boiling point of #100^@"C"#. In your case, the boiling point of the solution is said to be equal to #100.52^@"C"#, which means that the boiling-point elevation is equal to

#DeltaT_"b" = 100.52^@"C" - 100^@"C" = 0.52^@"C"#

Plug in your values to find the freezing-point depression

#DeltaT_"f" = (1.86 color(red)(cancel(color(black)(""^@"C kg mol"^(-1)))))/(0.512color(red)(cancel(color(black)(""^@"C kg mol"^(-1))))) * 0.52^@"C"#

#DeltaT_"f" = 1.89^@"C"#

Water has a normal freezing point of #0^@"C"#. The freezing-point depression is equal to #1.89^@"C"#, which means that the freezing point of the solution will be #1.89^@"C"# lower than the normal freezing point of pure water.

You thus have

#T_"f sol" = 0^@"C" - 1.89^@"C" = color(darkgreen)(ul(color(black)(-1.89^@"C")))#

SIDE NOTE The difference between this value and the value given to you in (a) is most likely caused by different values used for #K_"b"# and/or #K_"f"#.