Answer to Question #308885 in Chemistry for Junard

Question

Molality & Molarity 1.

Calculate the molality of a solution containing 10.8g of ethylene glycol (C2H6O2) in 360g of water.

Answer 1

Solution:

solute = ethylene glycol (C₂H₆O₂)

solvent = water (H₂O)

The molar mass of ethylene glycol (C₂H₆O₂) is 62.07 g/mol

Therefore,

Moles of C₂H₆O₂ = (10.8 g C₂H₆O₂) × (1 mol C₂H₆O₂ / 62.07 g C₂H₆O₂) = 0.174 mol C₂H₆O₂

Convert g to kg:

Kilograms of solvent = (360 g H₂O) × (1 kg / 1000 g) = 0.36 kg H₂O

Molality = Moles of solute / Kilograms of solvent

Therefore,

Molality of C₂H₆O₂ solution = Moles of C₂H₆O₂ / Kilograms of solvent

Molality of C₂H₆O₂ solution = (0.174 mol) / (0.36 kg) = 0.483 mol/kg = 0.483 m

Molality of C₂H₆O₂ solution = 0.483 m

Answer: The molality of C₂H₆O₂ solution is 0.483 m