Final answer:
To prepare 100.0 mL of a 15 M NaOH solution, 60.0 grams of solid NaOH are required, which is calculated using the molarity formula and the molar mass of NaOH (40.0 g/mol).
Step-by-step explanation:
To calculate the mass of NaOH needed for making a 100.0 mL of a 15 M NaOH solution from solid sodium hydroxide, we'll need to use the formula:
Mass of solute (g) = Molarity (M) × Volume of solution (L) × Molar mass of solute (g/mol)
First, we need to convert the volume from milliliters to liters since molarity is defined as moles of solute per liter of solution:
100.0 mL = 0.100 L
Next, we find the molar mass of NaOH using the atomic masses of its constituent elements:
Molar mass of NaOH = (1×22.990 g/mol) + (1×15.999 g/mol) + (1×1.008 g/mol) = 40.0 g/mol
Now we can calculate the mass of NaOH needed:
Mass of NaOH = 15 M × 0.100 L × 40.0 g/mol = 60.0 g
Therefore, 60.0 grams of solid NaOH are required to prepare 100.0 mL of a 15 M NaOH solution.