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5 August, 21:28

Calculate the mass of calcium carbonate present in a 50.00 mL sample of an

aqueous calcium carbonate standard, assuming the standard is known to

have a hardness of 75.0 ppm

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Answers (2)
  1. 5 August, 21:30
    0
    The mass of calcium carbonate in this sample = 3.74 * 10 ^-3 g

    Explanation:

    Mass of Calcium carbonate = Density of CaCO3 * Volume of the sample

    ⇒ CaCO3 has a density of 0.9977 g/mL

    ⇒ The sample has a volume of 50 mL

    Mass = 0.9977 g/mL * 50 mL = 49.885 g

    Since it has a hardness of 75 ppm = 75 parts per million = 75 mg/l = 7.5 * 10 ^-5 g / ml

    ⇒7.5 * 10^-5 * 48.885 g = 3.74 * 10 ^-3 g

    The mass of calcium carbonate in this sample = 3.74 * 10 ^-3 g
  2. 5 August, 21:33
    0
    The mass of calcium carbonate is 3,75 mg

    Explanation:

    Remember that when you work with ppm, you have mass solute / volume solution.

    In this case, your volume which is mL has to be in L.

    50 mL = 0,05L

    Ppm = mg / kg or mg/L

    75 mg/L = mass solute / 0,05L

    75 mg/L x 0,05L = 3,75 mg
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