Question

a 25.0 mL sample of sodium sulfate was analyzed by adding an excess of barium chloride to produce barium sulfate crystals

Na2SO4(aq) + BaCl2(aq) -> 2NaCl(aq) + BaSO4(s)

If 5.719g of BaSO4 was obtained what was the molarity of the original Na2SO4?

Answer 1

Answer on Question#49919 – Chemistry – Inorganic Chemistry

A 25.0 mL sample of sodium sulfate was analyzed by adding an excess of barium chloride to produce barium sulfate crystals

$$\mathrm{Na_2SO_4(aq)} + \mathrm{BaCl_2(aq)} \rightarrow 2\mathrm{NaCl(aq)} + \mathrm{BaSO_4(s)}$$

If 5.719g of BaSO₄ was obtained what was the molarity of the original Na₂SO₄?

Solution:

$$\nu = \frac{m}{M}; \quad \nu - \text{the mole (mol); } m - \text{the mass (g); } M - \text{the molar mass (g/mol);}$$ $$M(\mathrm{BaSO_4}) = 233\ \mathrm{g/mol}; \quad m(\mathrm{BaSO_4}) = 5.719\ \mathrm{g};$$ $$\nu(\mathrm{BaSO_4}) = 0.025\ \mathrm{mol};$$

According to the equation:

$$\nu(\mathrm{BaSO_4}) = \nu(\mathrm{Na_2SO_4}) = 0.025\ \mathrm{mol};$$ $$C = \frac{\nu}{V}; \quad C - \text{the molarity (mol/L); } V - \text{the volume of solution (L);}$$ $$C(\mathrm{Na_2SO_4}) = 1\ \mathrm{mol/L};$$

Answer: 1 mol/L.

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