Question

a) honey contains approximately 2.88 M fructose sugar (C6H12O6 MM=180.16 g/mol) and has a density of 1.36 g/mL. what is the boiling point of honey assuming it consists only of fuctose and water? kb=0.52 C/m for water

Answer 1

Molarity of fructose = 2.88 M

Assume the total volume is 1 liter.

Moles of fructose = 2.88

Molar mass of fructose = 180.16 g/mol

Mass of fructose = 2.88 * 180.16

= 518.86 g

Density of honey = 1.36 g/mL

Mass of honey = 1.36 * 1000

= 1360 g

Mass of water in honey = 1360 - 518.86

= 841.14 g = 0.841 kg

Molality of honey = moles of Fructose / Kilograms of water

= 2.88 / 0.841

= 3.42 m

Kb of water = 0.52 oC/m

i for fructose = 1

Increase in boiling point of water, T_b = i * Kb * m

= 1 * 0.52 * 3.42

= 1.78 oC

Boiling point of honey = Boiling point of water + T_b

= 100 + 1.78

Boiling point of honey = 101.78 oC