4) How many grams of NaCl must be added to 2.00 mL of water to make a 85.5 M solution? \( (1000 \mathrm{~mL}=1 \mathrm{~L}) \)

To prepare an 85.5 M solution, you can use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] First, convert 2.00 mL to liters: \[ 2.00 \, \text{mL} = 0.00200 \, \text{L} \] Next, solve for the moles of NaCl needed: \[ 85.5 \, \text{M} = \frac{\text{moles of NaCl}}{0.00200 \, \text{L}} \] \[ \text{moles of NaCl} = 85.5 \times 0.00200 = 0.171 \, \text{moles} \] Now, convert moles of NaCl to grams using its molar mass (approximately 58.44 g/mol): \[ \text{grams of NaCl} = 0.171 \, \text{moles} \times 58.44 \, \text{g/mol} \approx 10.00 \, \text{grams} \] So, you need to add approximately 10.00 grams of NaCl to 2.00 mL of water to make an 85.5 M solution.