Calculate the moles of aqueous calcium carbonate present in a 50.00mL sample of a calcium carbonate standard solution, assuming the standard is known to have a hardness of 125 ppm (hardness due to CaCO3). You can assume that the density of the sample is 1.000 g/mL and the molar mass of calcium carbonate is 100.0 g/mol.

1 ppm = 1 mg/L.

So 125 ppm = 125 mg/L.
You will have 125 mg/L x 0.050 L = 6.25 mg or 0.00625 g.
Then mols = grams/molar mass = ?

.0000625

To calculate the moles of aqueous calcium carbonate present in the given sample, you need to follow these steps:

Step 1: Calculate the mass of calcium carbonate (CaCO3) in the sample.
Given:
Volume of the sample (V) = 50.00 mL
Density of the sample = 1.000 g/mL

Mass = Density × Volume
Mass = 1.000 g/mL × 50.00 mL = 50.00 g

Step 2: Calculate the number of moles of calcium carbonate.
Given:
Molar mass of calcium carbonate (CaCO3) = 100.0 g/mol

Number of moles = Mass / Molar mass
Number of moles = 50.00 g / 100.0 g/mol = 0.500 mol

Therefore, there are 0.500 moles of aqueous calcium carbonate present in the 50.00 mL sample of the calcium carbonate standard solution.