Question

The mole fraction of an aqueous solution of magnesium sulfite is 0.29. Calculate the molarity (in mol/L) of the magnesium sulfite solution, if the density of the solution is 1.38 g mL^-1.

Determine the mole fraction of magnesium bromide in a 5.51 M aqueous solution of magnesium bromide. The density of the solution is 1.17 g mL^-1.

Answer 1

The mole fraction of an aqueous solution of magnesium sulfite is 0.29. Calculate the molarity (in mol/L) of the magnesium sulfite solution, if the density of the solution is 1.38 g mL^-1.

moles of magnesium sulfite = 0.29
moles of water = 1.00 - 0.29 =0.71

mass water = 0.71 x 18.02 g/mol=12.79 g
mass magnesium sulfite = 0.29 x 104.3682 g/mol = 30.27g

total mass of solution = 12.79+30.27= 43.06 g
volume solution =mass / density= 43.06 g/ 1.38 g/ml =31.20 mL = 0.0312 L

M = 0.29 / 0.0312=9.29 M

Determine the mole fraction of magnesium bromide in a 5.51 M aqueous solution of magnesium bromide. The density of the solution is 1.17 g mL^-1.

Let assume 1 L of solution

moles of magnesium bromide present
.. 1L solution x (5.51 mol magnesium bromide / L)

= 5.51 mol magnesium bromide

mass of magnesium bromide present
.. 1L solution x (5.51mol MgBr / L) x (184.113g MgBr / mol MgBr)

=1014.5 g MgBr

mass of solution =1L solution x (1000mL / 1L) x (1.17g solution / L solution)

=1170 g solution

then
.. mass H2O present = 1170 g solution - 1014.5 g MgBr

= 155.5 g H2O

then
.. moles H2O present = 155.5g H2O x (1 mol H2O / 18.02g H2O)

= 8.63 mol H2O

and finally
.. mole fraction MgBr = (5.51 mol magnesium bromide) / (5.51 mol magnesium bromide

+ 8.63mol H2O)

= 5.51/14.14

=0.39