Final answer:
To prepare a 400 ml of 1.20 M NaCl solution, 28.05 grams of NaCl are required, calculated by multiplying the moles of NaCl needed (0.48 moles) by its molar mass (58.44 g/mol).
Step-by-step explanation:
To prepare 400 ml of a 1.20 M solution of NaCl, you first need to determine the number of moles of NaCl required. The molarity (M) is defined as moles of solute per liter of solution. Thus, to find the moles of NaCl needed:
Molarity (M) = moles of solute / liters of solution
1.20 M = moles of NaCl / 0.4 L
Moles of NaCl = 1.20 M × 0.4 L = 0.48 moles
Next, use the molar mass of NaCl to find the mass in grams:
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl = moles × molar mass
Mass of NaCl = 0.48 moles × 58.44 g/mol = 28.05 grams
Therefore, you would need to add 28.05 grams of NaCl to 400 ml of water to prepare a 1.20 M solution.