Sofia A. answered 07/27/22
AP Chemistry Tutor with a PhD and College Teaching Experience
A brief Google search tells me that the purity of AR CaCO3 is about 99%.
The MW of CaCO3 is 100.0869 g/mol
Each 100.0869 g CaCO3 contain 40.078 g Ca.
1000 ppm Ca solution has concentration of 1000 mg Ca / L
500 mL of such solution should contain 500 mg Ca
The required mass of pure CaCO3 is then
500 mg Ca x (100.0869 g/mol CaCO3) / (40.078 g/mol Ca) = 1,249 mg CaCO3
A brief Google search tells me that the purity of AR CaCO3 is about 99%, so you need to weigh
1,249 mg / 0.99 = 1,261 mg