What volume (in liters) of a 0.60 mol/L CaCO3 solution is required to contain 105 g of CaCO3? (Molar mass of CaCO3 = 100 g/mol)

• A. 0.58 L
• B. 2.10 L
• C. 1.75 L
• D. 0.63 L

Answer: (C)

Verified step by step guidance

1. Identify the given information: concentration of CaCO3 solution is 0.60 mol/L, mass of CaCO3 required is 105 g, and molar mass of CaCO3 is 100 g/mol.
2. Convert the mass of CaCO3 to moles using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). So, calculate \(\text{moles of CaCO3} = \frac{105\ \text{g}}{100\ \text{g/mol}}\).
3. Use the concentration formula to relate moles and volume: \(C = \frac{n}{V}\), where \(C\) is concentration, \(n\) is moles, and \(V\) is volume in liters.
4. Rearrange the formula to solve for volume: \(V = \frac{n}{C}\). Substitute the moles calculated in step 2 and the concentration given to find the volume.
5. Calculate the volume using the values from previous steps to find the volume of CaCO3 solution required to contain 105 g of CaCO3.