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15 February, 21:03

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

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  1. 15 February, 22:09
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    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.
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