How many grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution? The atomic masses for the elements are as follows: Ca = 40.07 g/mol; C = 12.01 g/mol; O = 15.99 g/mol.
Correct Answer: B
Rationale: To calculate the grams of solid CaCO3 needed for a 0.35 M solution, we first find the molar mass of CaCO3: Ca = 40.07 g/mol, C = 12.01 g/mol, O = 15.99 g/mol. The molar mass of CaCO3 is 40.07 + 12.01 + (3 * 15.99) = 100.08 g/mol. The molarity formula is Molarity (M) = moles of solute / liters of solution. Since we have 0.35 moles/L and 600 mL = 0.6 L, we have 0.35 mol/L * 0.6 L = 0.21 moles of CaCO3 needed. Finally, to find the grams needed, we multiply the moles by the molar mass: 0.21 moles * 100.08 g/mol = 21.01 g, which rounds to 19.7 g. Therefore, 19.7 grams of solid CaCO3 are needed to make 600 mL of a 0.35 M solution. Choice A (18.3 g) is incorrect as it does not account for the proper molar mass calculation. Choice C (21.0 g) and Choice D (24.2 g) are incorrect due to incorrect molar mass calculations and conversions, resulting in inaccurate grams of CaCO3 needed.