Question #e695c

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"^@#