Molarity (M) = moles of solute (mol) / Volume of the solution (L)
Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL
0.25 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 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
Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.
