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25 June, 09:16

Express the concentration of a 0.0390 M aqueous solution of fluoride, F -, in mass percentage and in parts per million (ppm). Assume the density of the solution is 1.00 g/mL.

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  1. 25 June, 09:23
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    Mass percentage → 0.074 %

    [F⁻] = 741 ppm

    Explanation:

    Aqueous solution of flouride → [F⁻] = 0.0390 M

    It means that in 1L of solution, we have 0.0390 moles of F⁻

    We need the mass of solution and the mass of 0.0390 moles of F⁻

    Mass of solution can be determined by density:

    1g/mL = Mass of solution / 1000 mL

    Note: 1L = 1000mL

    Mass of solution: 1000 g

    Moles of F⁻ → 0.0390 moles. 19g / 1 mol = 0.741 g

    Mass percentage → (Mass of solute / Mass of solution). 100

    (0.741 g / 1000 g). 100 = 0.074 %

    Ppm = mass of solute. 10⁶ / mass of solution (mg/kg)

    0.741 g. 1000 mg/1g = 741 mg

    1000 g. 1 kg/1000 g = 1kg

    741 mg/1kg = 741 ppm
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