Eeek, chemistry is hard?

Question

# 103)

A glucose solution contains 55.8 g of glucose (C6H12O6) in 455 g water. compute the freezing and boiling point of the solution. (Assume a density of 1.00 g/mL for water)

WOW I'm confused... Please explain so I can understand for future times this question comes up. thanks :)

Answer 1

theres a little more to thhe question than just the  stuff you've given.

did you get given a number to do with the concentration of glucose in solution vs BP, or the like?

as a note BP and MP depend on how strong the intermolecular, in this case the bonds between water molecules, are. by adding a substance you change the strength of these bonds. for example salt will lower yor MP as it affects the hydrogen bonds in between adjacent water molecules and allows them to break at a lower energy. it also, as an ionic substance, causes the BP to increase as it forms ionic intermolecular bonds between the water, which means more energy is required to boil the water, and so a higher heat.

Answer 2

55.8 g of glucose = 0.310 mol (55.8 g / 180.15 g/mol)

molality = moles / kg solvent = 0.310 / 0.455 = 0.681 molal

fp depression = 0.681 molal x 1.86 deg/molal = 1.27

fp = -1.27 deg C

bp elevation = 0.681 molal x 0.51 deg/molal = 0.35

bp = 100.35 deg C at 760 mm Hg

The solvent (water) has a characteristic molal freezing point depression constant (1.86 deg/molal) and a characteristic molal boiling point elevation constant (0.51 deg/molal).