Final answer:
To prepare 5.7 L of a 1.5 M NaOH solution, one would need 342 grams of NaOH, calculated by multiplying the required moles (1.5 M × 5.7 L) by the molar mass of NaOH (40 g/mol).
Step-by-step explanation:
To calculate the mass of NaOH needed to prepare 5.7 L of a 1.5 M NaOH solution, one must use the molarity formula: M = moles/L. First, we need to calculate the moles of NaOH required and then convert the moles to grams using the molar mass of NaOH.
Step 1: Calculate moles of NaOH needed
Moles = Molarity × Volume (L)
Moles = 1.5 mol/L × 5.7 L = 8.55 mol
Step 2: Convert moles to grams
Grams = Moles × Molar mass of NaOH
NaOH has a molar mass of approximately 40 g/mol, so:
Grams = 8.55 mol × 40 g/mol = 342 grams
Therefore, 342 grams of NaOH are needed to prepare a 5.7 L of a 1.5 M NaOH solution.