To prepare **1000 mL (1.0 L) of a 1.0 M NaCl solution**: **Formula:** \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \] So, \[ \text{Moles of NaCl needed} = 1.0~\text{mol/L} \times 1.0~\text{L} = 1.0~\text{mol} \] \[ \text{Mass (g)} = \text{moles} \times \text{molar mass} \] \[ \text{Mass (g)} = 1.0~\text{mol} \times 58.44~\text{g/mol} = \boxed{58.44~\text{g}} \] **Answer:** **58.44 grams** of NaCl are needed to prepare 1000 mL of a 1.0 M NaCl solution.