Final answer:
To prepare 6.5 L of a 0.270 M CaCl2 solution, the chemist should use 194.92 grams of CaCl2, calculating moles required from the given molarity and volume, and converting it to grams using CaCl2's molar mass.
Step-by-step explanation:
To calculate the mass of CaCl2 the chemist should use to make 6.5 L of a 0.270 M CaCl2 solution, we need to use the formula for molarity (M), which is moles of solute divided by the volume of solution in liters (M = moles/L). First, we calculate the moles of CaCl2 needed:
Moles of CaCl2 = Molarity (M) × Volume (L) = 0.270 M × 6.5 L = 1.755 moles
Next, the molar mass of CaCl2, which is approximately 110.98 g/mol, is used to convert moles to grams:
Mass of CaCl2 = Moles × Molar Mass = 1.755 moles × 110.98 g/mol = 194.92 g
Therefore, the chemist should use 194.92 grams of CaCl2 to prepare 6.5 L of a 0.270 M solution.