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

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.