Final answer:
To prepare a 0.25 M CaCO3 solution, one must calculate the mass of calcium carbonate corresponding to 0.25 moles, but in practice, this exceeds its solubility in water. Therefore, a complete dissolution in 100 ml is not feasible, leading to a saturated solution.
Step-by-step explanation:
To prepare a 0.25 molar (0.25 M) calcium carbonate (CaCO3) standard solution, you first need to calculate the mass of CaCO3 that corresponds to 0.25 moles. Since the molar mass of CaCO3 is approximately 100.09 g/mol, we multiply 0.25 moles by this molar mass to get the mass required: 0.25 moles × 100.09 g/mol = 25.0225 grams.
Next, you would usually dissolve this mass in enough water to make a total volume of one liter for a standard molar solution. However, due to the insolubility of calcium carbonate in water, you will not be able to dissolve 0.25 moles in 100 milliliters of water, as it exceeds the solubility limit. Instead, saturation will occur, with excess undissolved CaCO3 remaining in the container.
For a truly homogeneous solution, you must ensure the volume of solvent allows for complete dissolution of the solute or adjust the amount of solute accordingly. It's essential to mix the solution thoroughly until the solid is completely dissolved or until no more solute can be dissolved, indicating the solution is saturated. Accurately measuring both solute and solvent weight is crucial for precise solution preparation.