# Question #8be0e

##### 1 Answer

Boiling point:

Freezing point:

#### Explanation:

To solve this problem, you need to know the value of water's *cryoscopic* and *ebullioscopic constants*,

#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

*freezing-point depression*;

*van't Hoff factor*, equal to

and

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

*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,

#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,

#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")#