Determine the molarity of sodium dichromate and the mole fraction of ammonium nitrite.

Question

1. The mole fraction of an aqueous solution of sodium dichromate is 0.283. Calculate the molarity (in mol/L) of the sodium dichromate solution, if the density of the solution is 1.50 g mL^-1.
2. Determine the mole fraction of ammonium nitrite in a 10.69 M aqueous solution of ammonium nitrite. The density of the solution is 1.13 g mL^-1.

Answer 1

For these problems we will use the definition of morality to help solve the problem. MOLARITY = moles solute per liter of solution. So we must determine moles of solute and then liters of solution.

1).

Moles sodium dichromate: let’s assume total moles (solute + solvent) = 1. Then since the mole fraction of the sodium dichromate is 0.283, that means we have 0.283 moles of the solute. This also means we have 1 - 0.283 = 0.717 moles of water.

Volume of solution: we will use the density and the mass to find the volume.

mass sodium dichromate = 0.283 mol x 262 g/mol = 74.1 g

mass H2O = 0.717 mol x 18 g/mol = 12.9 g

Total mass = 87.0 g

Volume of soln = 87.0 g x 1 ml/1.50 g = 58.0 ml = 0.058 L

MOLARITY = 0.283 mols/0.058 L = 4.88 M

Be sure to check all math.

Do part (2) the same way.