Question

In a CaF₂ solution containing 23.9g CaF₂ in 500 mL aqueous solution, what is the molarity and molality of the F-ions? (assume a density of 1.00 g/mL)

Answer 1

molarity

molarity is a concentration unit, defined to be the number of moles of solute divided by the number of liters of solution.

Molarity = Number of moles of solute / Liters of solution

number of moles of solute CaF₂ = Weight of CaF₂ / Molececular mass of CaF₂

given weight of CaF₂ = 23.9 g

Molecular mass of CaF2 = (atomic mass of Ca) + (2 x atomic mass of F) = 40.078 + (2 x 18.998) = 78.074 g/mol

number of moles of CaF₂ = 23.9 / 78.074 = 0.306 moles

given liters of solution = 500 mL = 0.5 L

molarity = Number of moles of CaF₂ / Liters of solution = 0.306 / 0.5 = 0.612 moles per liter

Molality

molality of given solution is defined as the total number of moles of solute per kilograms of solvent present in the solution.

number of moles of CaF₂ = 0.306 moles

density of solvent = 1 g/mL = 1 Kg/L

volume of solvent = 0.5 L

mass of solvent = volume of solvent x density of solvent = 0.5 L x 1 Kg/L = 0.5 Kg

molality = number of moles of CaF₂ / Mass of solvent in Kg = 0.306 / 0.5 = 0.612 moles per kilogram