# An excess of barium nitrate was added to an aqueous solution of sodium chromate weighing 100 g.

**An excess of barium nitrate was added to an aqueous solution of sodium chromate weighing 100 g. This resulted in a precipitate weighing 10.12 g. Determine the mass fraction of sodium chromate in the original solution. The answer should be: 0.0648, or 6.48%**

Let’s find the received amount of sodium chromate.

M BaCrO4 = 137 +52 + 16 x 4 = 253 grams / mol;

N BaCrO4 = 10.12 / 253 = 0.04 mol;

To obtain this amount of insoluble precipitate, it is necessary to take the same amount of sodium chromate. Let’s calculate its weight.

For this purpose, we multiply the weight of the precipitate by its molar mass, which is equal to the sum of the molar weights of the atoms included in the molecule.

M Na2CrO4 = 23 x 2 +52 + 16 x 4 = 162 grams / mol;

m Na2CrO4 = 0.04 x 162 = 6.48 grams;

The mass fraction of sodium chromate in the initial solution will be 6.48 / 100 = 0.0648 = 6.48%;