Calculate the concentration of sulfate in a solution made from 65.0 mL of 0.55 M sodium sulfate and 85.0 mL of 1.25 M iron(III) sulfate?

we are given the molarity and the volume of solution. The only issue is that the volume is given in mL instead of L. This issue can be fixed by using the following conversion factor:

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1000

m

L

=

1

L

Therefore, if we divide 50mL by 1000mL we will obtain a value of 0.05L.

Next, the equation has to be rearranged to solve for the moles of solute:

Moles of solute = Molarity

×

Liters of solution

Now, multiply 0.360 M by 0.05:

0.360 mol

1 L

×

0.05 L

=

0.018 mol

To obtain the mass of solute, we will need to the molar mass of NaCl, which is 58.44 g/mol:

Finally, multiply the number of moles by 58.44 g/mol

0.018

mol

×

58.44

g

1

mol

Boom, here it is:

1.05

g