Chemistry Help! Molality!

What weight of dextrose, C6H12O6, dissolved in 2000 grams of water, will raise the boiling-point to 102.04C at one atmosphere pressure?

ΔT = (Kb) m

where

ΔT = change in boiling point of the solvent
Kb = molal boiling point elevation constant = 0.512 °C / m
m = molality = moles of solute / kg of solvent


Let's first calculate the molality:

m = moles of solute / kg of solvent
m = (mass of solute / molecular weight of solute) / kg of solvent
m = (x grams of dextrose / 180.16 g/mol) / 2 kg of water


ΔT = (Kb) m

2.04 = (0.512) [(x / 180.16) / 2]

2.04 = (0.256) (x / 180.16)

x = (2.04)(180.16) / (0.256)

x = 1435.65 grams of dextrose



Hope this helps...good luck!