Answer:
Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL = 0.25 L
moles in 250.0 mL = molarity x volume of the solution
= 3.000 M x 0.25 L
= 0.75 mol
Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.
Moles (mol) = mass (g) / molar mass (g/mol)
Moles of NaCl in 250.0 mL = 0.75 mol
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl in 250.0 mL = Moles x Molar mass
= 0.75 mol x 58.44 g/mol
= 43.83 g
Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.
Explanation:
