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5 years ago
1- A solution is 6.53% K2SO4. An experiment require 5.54g K2SO4. How many grams of the solution do you need for this experiment?

2- An aqueous solution made from 0.758g of barium chloride, BaCl2. If the volume of the solution is 52.80mL, What is the molarity of barium chloride in the solution?

3- How many milliliters of 0.130 M  Na3PO4 are required to give 0.0165 mol of sodium phosphate, Na3PO4?


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5 years ago
1.
\(\frac{100 g solution}{6.53 g {{K}_{2}}{{SO}_{4}}}*5.54 g {{K}_{2}}{{SO}_{4}}\) = 84.8 g solution

2.
\(\frac{0.758 g {{BaCl}_{2}}}{52.80 mL {{BaCl}_{2}}}*\frac{1 mol {{BaCl}_{2}}}{208.2324 g {{BaCl}_{2}}}*\frac{1000 mL}{1 L}\) = 0.0689 M BaCl2 solution

3.
\(\frac{1 L {{Na}_{3}}{{PO}_{4}}}{0.130 mol {{Na}_{3}}{{PO}_{4}}}*0.0165 mol {{Na}_{3}}{{PO}_{4}}*\frac{1000 mL}{1 L}\) = 127 mL Na3PO4 solution
Pretty fly for a SciGuy
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