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A solution is 12.5% Na\(_3\)PO\(_4\) by mass. How many grams of Na\(_3\)PO\(_4\) are there in a 35.9 mL solution of Na\(_3\)PO\(_4\), if the density of the solution is 1.10 g/mL?

Answer :

Answer: 4.94 g

Explanation:

Mass of solution of Na₃PO₄ = volume × density

= 35.9 mL × 1.10 g/mL

= 39.49 g

Now, since the solution only contains 12.5% Na₃PO₄, then the mass of Na₃PO₄ is 12.5% of 39.49 g

12.5% × 39.49 g = 4.94 g

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