Asolution contains an unknown mass of dissolved barium ions. when sodium sulfate is added to the solution, a white precipitate forms. the precipitate is filtered and dried and then found to have a mass of 236 mg. what mass of barium was in the original solution? (assume that all of the barium was precipitated out of solution by the reaction.)

0.1388 g

Explanation:

The mass of obtained on precipitation = 236 mg

1 mg = 0.001 g

Thus, Mass of = 0.236 g

Molar mass of = 233.43 g/mol

The formula for the calculation of moles is shown below:

Thus,

Moles of = 0.001011 moles

According to the reaction,

Thus, moles of barium = 0.001011 moles

Molar mass of barium = 137.327 g/mol

Thus, Mass = Moles * Molar mass = 0.001011*137.327 g = 0.1388 g

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