9.00 L of a solution that is 60.0 ppm in Na+, starting with solid Na2SO4

Question

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1 point
Dissolve 1.7 g Na2SO4 in enough water to give a final volume of 9.00 L.
Dissolve 18 g Na2SO4 in enough water to give a final volume of 9.00 L.
Dissolve 2.7 g Na2SO4 in enough water to give a final volume of 9.00 L

Answer 1

To determine how much Na2SO4 to dissolve in order to make a 60.0 ppm Na+ solution in a final volume of 9.00 L, we first need to calculate the molecular weight of Na2SO4:

2(Na) + 1(S) + 4(O) = 2(22.99) + 32.07 + 4(16.00) = 45.98 + 32.07 + 64.00 = 142.05 g/mol

Next, we need to calculate the molar concentration of Na+ ions in a 60.0 ppm solution:

60.0 ppm = 60.0 mg/L = 60.0 mg/L / 1,000 mg/g = 0.060 g/L

The atomic weight of Na is 22.99 g/mol, so the molarity of Na+ ions in a 60.0 ppm solution is:

(0.060 g/L) / (22.99 g/mol) = 0.00261 mol/L

Since there are 2 sodium ions in 1 mole of Na2SO4, the molarity of Na2SO4 is twice the molarity of Na+ ions.

0.00261 mol/L * 2 = 0.00522 mol/L

Finally, we can calculate how much Na2SO4 to dissolve in 9.00 L of solution to achieve a molarity of 0.00522 mol/L:

0.00522 mol/L * 142.05 g/mol = 0.741 g/L

0.741 g/L * 9.00 L = 6.67 g

Therefore, dissolve 6.67 g of Na2SO4 in enough water to make a final volume of 9.00 L to achieve a 60.0 ppm Na+ solution.