A 20 g mixture containing copper sulfate and sodium chloride was dissolved in water. Then an excess of barium chloride solution was added to the mixture, which resulted in a precipitate weighing 9.32 g. Determine the mass fraction of sodium chloride in the initial mixture.
Let’s calculate the chemical amount of synthesized barium sulfate.
For this purpose, we divide its weight by its molar mass, equal to the sum of the molar weights of the atoms included in the molecule.
M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;
N BaSO4 = 9.32 / 233 = 0.04 mol;
To obtain such an amount of sediment in the amount of 0.04 mol, it is necessary to take the same amount of copper sulfate.
Let’s determine its weight.
M CuSO4 = 64 + 32 + 16 x 4 = 160 grams;
The weight of 0.04 mol of copper sulfate will be equal to: equal to:
m CuSO4 = 160 x 0.04 = 6.4 grams;
The weight of sodium chloride in the initial mixture will be 20 – 6.4 = 13.6 grams;
The mass fraction of sodium chloride is 13.6 / 20 = 0.68 = 68%;
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