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6 November, 04:15

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate is filtered and dried and found to have a mass of 258 mg. What mass of barium was in the original solution assume that all of the barium was precipitated out of the solution you the reaction.

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  1. 6 November, 04:20
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    The mass of barium in the original solution was 152mg.

    Explanation:

    The white precipiate that is formed when sodium sulfate, Na₂SO₄, reacts with barium ions in solution, Ba²⁺, is barium sulfate, BaSO₄.

    The molar mass of BaSO₄ is 233.38 g/mol. Thus, the number of moles of BaSO₄ is:

    #moles = mass in grams / molar mass #moles = 258mg * (1g/1,000mg) * (1mol/233.38g) = 0.00110549mol

    Since 1 mole of BaSO₄ contains 1 mole of Ba, there are 0.00110549 mol of Ba²⁺ ions.

    Use the atomic mass of Ba to calculate the mass in grams in the original solution:

    mass = number of moles * atomic mass mass = 0.00110549mol * 137.327g/mol = 0.1518g. mass = 151.8mg = 152mg

    Then, with 3 significant figures, there were 152mg of barium in the original solution.
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