How many grams of nacl are required to make 250.0 ml of a 3.000 m solution?

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 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.

A) the enthralling of formation for pure element in its standard state is always positivehope this : )