Question

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?

Answer 1

Remember the vital equations for this:
(freezing point depression) =  (Depression Constant) * (molality of solution)
Molality = (moles of solute) / (kg of solvent)
Step 1: Find molar mass of glucose
6 carbon * (12.01 g/mol) + 12 hydrogen * (1.01 g/mol) + 6 oxygen (16.00 g/mol)
= 180.18 g/mol
21.5 g glucose * (1 mole glucose / 180.18 g glucose) = 0.119 moles of glucose
Step 2: Convert grams of solvent to kilograms of solvent
255 g H2O * (1 kg/1000g) = 0.255 kg H2O
Step 3: Find the molality of the solution
molality = (moles solute)/(kg solvent)
molality = (0.119 moles glucose)/(0.255 kg H2O solvent) = 0.468 m (molal)
Step 4: Plug into depression equation
FP-depression = (-1.86°C/m) * (0.468 m) = -0.870 °C