How much NaCl is contained in 330 mL of a 0.205 M solution of NaCl?

The answer to your question is 3.96 grams

Explanation:

Data

mass of NaCl = ?

volume = 330 ml = 0.330 L

Molarity = 0.205

Process

1.- Calculate the moles of NaCl in solution

Molarity = moles / volume

moles = Molarity x volume

-Substitution

moles = 0.205 x 0.330              

moles = 0.0677

2.- Calculate the molar mass of NaCl

Molar mass = 23 + 35.5

                   = 58.5 g

3.- Use proportions to calculate the mass of NaCl in solution

                      58.5 g of NaCl 1 mol

                        x                        0.0677 moles

                        x = (0.0677 x 58.5) / 1

                        x = 3.96 / 1

                        x = 3.96 grams

4.- Conclusion

There are 3.96 grams of NaCl in 330 ml of 0.205 M solution.

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

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