A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.
-
ΔTf = molality * Kf * i
where:
ΔTf = freezing point depression = unknown
molality = (mass of solute / molar mass of solute) / kg of solvent
molality = (21.5 g / 180 g/mol) / (255/1000)kg
molality = 0.4684 mole / kg
Kf = cryoscopic constant = 1.86 °C kg/mol
i = Van't Hoff factor = 1 (for all non-electrolyte; glucose is non-electrolyte)
ΔTf = (0.4684 mole/kg)(1.86 °C kg/mol)(1)
ΔTf = 0.8712 °C ====> answer
where:
ΔTf = freezing point depression = unknown
molality = (mass of solute / molar mass of solute) / kg of solvent
molality = (21.5 g / 180 g/mol) / (255/1000)kg
molality = 0.4684 mole / kg
Kf = cryoscopic constant = 1.86 °C kg/mol
i = Van't Hoff factor = 1 (for all non-electrolyte; glucose is non-electrolyte)
ΔTf = (0.4684 mole/kg)(1.86 °C kg/mol)(1)
ΔTf = 0.8712 °C ====> answer