Question

A solution is made by dissolving 3.35 g of fructose in 35.0 ml of water. What is the molality of fructose in the solution? Assume density of water is 1.00 g/ml.

Answer 1

Answer:

m = 0.531 molal

Explanation:

∴ m fructose = 3.35 g

∴ V water = 35.0 mL

∴ ρ H2O = 1 g/mL

• molality = moles solute / Kg solvent

∴ Mw fructose = 180.16 g/mol

⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose

⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O

⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O

⇒ m = 0.531 molal