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Question

A solution contains 25 ppm of NaNO₃ by mass in water. What is its molarity if the density is 1 g/mL? (Molar mass: NaNO₃ = 85 g/mol)

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Explanation

25 ppm = 25 g NaNO₃ in 10⁶ g solution. Volume = 1000 L. Moles = 25/85 ≈ 0.2941 mol; molarity = 0.2941/1000 ≈ 2.941 × 10⁻⁴ M.

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