Question

HELP MEDAL WILL BE AWARDED

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 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.

Answer 1

Strategy:
1.Step 1: Calculate the freezing point depression of benzene.

deltaTf = (Freezing point of pure solvent) - (Freezing point of solution)
(5.5 oC) - (2.8 oC) = 2.7 oC


2.Step 2 : Calculate the molal concentration of the solution.

molality = moles of solute / kg of solvent
moles of naphthalene = (1.60 g) (1 mol / 128 g) = 0.0125 mol naphthalene
molality of solution = (0.0125 mol) / (0.0200 kg) = 0.625 m


3.Step 3: Calculate Kf of the solution.

deltaTf = (Kf) (m)

(2.7 oC) = (Kf) (0.625 m)

Kf = 4.3 oC/m

Answer 2

A solution is made by dissolving 2.5 moles of sodium chloride (NaCl) in 198 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem.

Answer 3

what about this one ^ @shadowhuntergaming im not sure what the molal boiling point constant is