Question

A solution is made by dissolving 89.0 g MgCl2 in enough water to make a 400.0 mL solution. If the density of the solution is 1.52 g/mL:

a. What is the molality of the solution? (m MgCl2)

b. What is the molarity of the solution? (M MgCl2)

c. What is the mass % of MgCl2 in the solution?

d. What is the mol fraction of MgCl2 in the solution?

Answer 1

mass of solute = 89.0 g  

mass of solution = 1.52 x 400 = 608 g

mass of solvent = 608 - 89 = 519 g

a) molality = (W /MW) (1000 / mass of solvent in g)

molality = (89 / 95.2) (1000 / 519)

molality = 1.80 m

b) molarity = (W /MW) (1000 / V in mL)

molarity = (89 / 95.2) (1000 / 400)

molarity = 2.34 M

c) % mass = (mass of solute / mass of solution) x 100

% mass = (89 / 608) x 100

% mass = 14.64 %

d ) moles of MgCl2 = 89 / 95.2 = 0.935

moles of H2O = 519 / 18 = 28.83

total moles = 28.83 + 0.935 = 29.765

X_MgCl2 = moles of MgCl2 / total moles

X_MgCl2 = 0.935 / 29.765

X_MgCl2 = 0.031