How many grams of NaCl is needed to make 750mL of a 2.4 M NaCl solution?

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:

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).

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