Final answer:
To prepare 100 mL of a 10,000 μg/dL standard solution of Ca²⁺, 0.1 grams of CaCO₃ are required, based on the molar mass of CaCO₃ and the conversion from μg/dL to g/L.
Step-by-step explanation:
To calculate the amount of CaCO₃ needed to prepare 100 mL of a 10,000 μg/dL standard solution of Ca²⁺, we first need to convert the concentration to a more useful unit (grams per liter). Since 10,000 μg/dL is equivalent to 100 mg/dL, we have 100 mg/dL * 10 dL/L = 1000 mg/L or 1 g/L for our solution concentration.
Now, the molar mass of CaCO₃ (calcium carbonate) is 100.09 g/mol. To find out how many grams are in 100 mL, we use the formula:
grams = concentration (g/L) × volume (L)
So for 100 mL (which is 0.1 L), we have:
grams of CaCO₃ = 1 g/L × 0.1 L = 0.1 g
Therefore, 0.1 grams of CaCO₃ are required to prepare 100 mL of a 10,000 μg/dL standard solution of Ca²⁺.