Question #8be0e

1 Answer
Sep 20, 2015

Answer:

Boiling point: #-1.4""^@"C"#
Freezing point: #100.4""^@"C"#

Explanation:

To solve this problem, you need to know the value of water's cryoscopic and ebullioscopic constants, #K_f# and #K_b#, respectively.

#K_f = 1.853""^@"C kg mol"""^(-1) " "#

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

The two equations that you will use to get the boling point and freezing point on the solution are

#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;
#b# - the molality of the solution.

and

#DeltaT_b = 8 * K_b * b" "#, where

#DeltaT_b# - the boiling-point elevation.

The molality of the solution is defined as the number of moles of solute, in your case fructose, divided by the mass of the solvent, expressed in kilograms.

Use fructose's molar mass to determine the number of moles your solution contains

#20color(red)(cancel(color(black)("g"))) * "1 mole"/(180.16color(red)(cancel(color(black)("g")))) = "0.111 moles fructose"#

The molality of the solution will thus be

#b = n_"fructose"/m_"water"#

#b = "0.111 moles"/(150 * 10^(-3)"kg") = 0.74"mol"/"kg"#

The freezing-point depression will be

#DeltaT_f = 1 * 1.853""^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.74color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("kg"))) = 1.37""^@"C"#

The freezing point of the solution, #T_f#, will be

#DeltaT_f = T_f^@ - T_f implies T_f = T_f^@ - DeltaT_f#

#T_f = 0""^@"C" - 1.37""^@"C" = color(green)(-1.4""^@"C")#

The boiling-point elevation will be

#DeltaT_b = 1 * 0.512""^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.74color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("kg"))) = 0.38""^@"C"#

The boiling point of the solution, #T_b#, will be

#DeltaT_b = T_b - T_b^@ implies T_b = DeltaT_b + T_b^@#

#T_b = 0.38""^@"C" + 100""^@"C" = color(green)(100.4""^@"C")#