Question

A solution has a 0.102 mole fraction of CaCl₂ (MM 110.98 g/mol) in water (MM 18.02 g/mol). The density of the solution is 1.05 g/mL. What is the molality, mass percent and molarity of CaCl₂ in the solution?

Answer 1

If we have a mole fraction of CaCl₂ x = 0.102 this means we have 0.102 moles of CaCl₂ per every 0.898 moles of H₂O.

a) Molality of the solution: to calculate molality we use the formula

m = n solute / kg solvent

Where n solute = 0.102 moles of CaCl₂ and we need to convert those 0.898 moles of H₂O to kilograms using the molar mass of water (18.02 g/mol).

kg solvent = 0.898 moles H₂O (18.02 g H₂O / 1 mol H₂O) (1kg H₂O / 1000g H₂O)

kg solvent = 0.0162 kg

Substitue n solute and kg solvent in the formula:

m = 0.102 mol / 0.0162 kg

m = 6.30 mol/kg

b) Mass percent: to calculate mass percent we use the formula:

%m/m = mass of solute/mass of solution x 100

We need to convert those 0.102 moles of CaCl₂ to grams (mass of solute) and we already know the mass of the solvent (0.0162kg = 16.2g) so the mass of solution is the sum of the mass of solute + mass of solvent.

mass of solute = 0.102 mol CaCl₂ (110.98g CaCl₂ / 1mol CaCl₂)

mass of solute = 11.32g CaCl₂

mass of solution = mass of solute + mass of solvent

mass of solution = 11.32g + 16.2g

mass of solution = 27.52g

Substitute in the formula:

%m/m = 11.32g / 27.52g x 100

%m/m = 41.2%

c) Molarity of the solution: to calculate molarity use the formula:

M = n solute / Liters of solution

We already have the n solute (0.102 mol CaCl₂) but we need to convert the mass of the solution to mL using density and convert those mL to L using the conversion factor. We already got the mass of the solution in the previous step (27.52g).

L of solution = 27.52g (1 mL / 1.05g) (1L / 1000mL)

L of solution = 0.0262L

Finally, replace the values in the formula:

M = 0.102mol / 0.0262L

M = 3.89 mol/L