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28 May, 16:11

A sample of tap water contains 1.95 x 10^3 ppm caco3. assuming the density of the tap water is 1.00 g/ml, calculate the mass of calcium carbonate in 2.0 l of water

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  1. 28 May, 19:20
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    3.9 grams CaCO3

    The mass of 2.0 L of water with a density of 1.00 g/ml is 2000 grams. And 1 ppm of that is 2000 / 1000000 = 0.002 grams. So just multiply by the ppm of CaCO3, giving 0.002 g * 1.95x10^3 = 3.90 grams.

    Since the least accurate datum we have is 2 significant figures, the result should be rounded to 2 significant figures, giving 3.9 grams.
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