When 20.0 mL of a 0.250 M (NH₄)₂S solution is added to 150.0 mL of a solution of Cu(NO₃)₂, a CuS precipitate forms.The precipitate is then filtered from the solution, dried, and weighed.If the recovered CuS is found to have a mass of 0.3491 g, what was the concentration of copper ions in the original Cu(NO₃)₂ solution? A)3.65 × 10^-3 M B)1.22 × 10^-2 M C)3.33 × 10^-2 M D)4.87 × 10^-2 M E)2.43 × 10^-2 M

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Quizplus · Accepted Answer

The mass of CuS can be used to find the moles of CuS, which equals the moles of Cu²⁺ initially present. The molar mass of CuS is approximately 95.61 g/mol. Moles of CuS = 0.3491 g / 95.61 g/mol = 0.00365 mol. The total volume of the solution is 170.0 mL (20.0 mL + 150.0 mL). The concentration of Cu²⁺ is 0.00365 mol / 0.150 L = 0.0243 M.

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