Question #e695c

1 Answer
Nov 11, 2015

Answer:

#T_"f" = -1.28^@"C"#

Explanation:

I'm not entirely sure why the problem provides the density of the water, since that is not needed to find the boiling point and the freezing point of the solution.

I can only assume that your solution contains #"57.8 g"# glucose dissolved in #"465 mL"# of water. Since the density of water is to be taken as #"1.00 g/mL"#, this will not make any difference for the final result.

So, the first thing would be to determine the mass of water

#465color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "465 g"#

The equation for freezing-point depression looks like this

#DeltaT_f = i * K_f * b" "#, where

#DeltaT_f# - the freezing-point depression;
#i# - the van't Hoff factor, equal to #1# for non-electrolytes;
#K_f# - the cryoscopic constant of the solvent;
#b# - the molality of the solution.

In your case, glucose is a non-electrolyte, which means that it does not dissociate to form ions in solution. As a result, you have #i=1#.

In order to calculate the molality of the solution, you need to know two things

  • the number of moles of solute, in your case sucrose
  • the mass of the solvent, in your case water, expressed in kilograms

Use glucose's molar mass to determine how many moles you have in that sample

#57.8color(red)(cancel(color(black)("g"))) * " 1 mole glucose"/(180.156color(red)(cancel(color(black)("mL")))) = "0.3208 moles glucose"#

The molality of the solution will thus be - do not forget to convert the mass of the water from grams to kilograms

#color(blue)(b = n_"solute"/m_"solvent")#

#b = "0.3028 moles"/(465 * 10^(-3)"kg") = "0.6899 moles/kg" = "0.6899 molal"#

You can find the values of water's cryoscopic and ebullioscopic constants here

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

So, plug in your values and determine the freezing-point depression

#DeltaT_"f" = 1 * 1.86^@"C" color(red)(cancel(color(black)("kg")))/color(red)(cancel(color(black)("mol"))) * 0.6899color(red)(cancel(color(black)("moles")))/color(red)(cancel(color(black)("kg"))) = 1.283^@"C"#

The freezing-point depression is defined as

#DeltaT_"f" = T_"f"^@ - T_"f"" "#, where

#T_"f"^@# - the freezing point of the pure solvent
#T_"f"# - the freezing point of the solution

This means that the freezing point of the solution will be

#T_"f" = T_"f"^@ - DeltaT_"f"#

#T_"f" = 0^@"C" - 1.283^@"C" = color(green)(-1.28^@"C")#

I'll leave the calculation of the solution's boiling point to you as practice. The equation for boiling-point elevation is

#DeltaT_b = i * K_b * b" "#, where

#DeltaT_b# - the boiling-point elevation;
#i# - the van't Hoff factor
#K_b# - the ebullioscopic constant of the solvent;
#b# - the molality of the solution.

Plug in your values and solve for #DeltaT_b#, then use

#DeltaT_"b" = T_"b" - T_"b"^@#