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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Ответ:
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
Ответ:
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
Ответ:
Depends like exersise?
Explanation: