A seawater sample has the following composition: Mass Concentrations Sodium 10,7

Question

A seawater sample has the following composition: Mass Concentrations Sodium 10,700 ppm (mg/kg) Potassium 399 ppm Calcium 412 ppm Magnesium 1294 ppm Barium 25 ppb (mu g/kg) Chloride 19,350 ppm Sulfate 2,700 ppm-S04 Bicarbonate 145 ppm-HCo3 Silicate Si(OH)4 degree 8 ppm- Si(OH)degree 4 Molar Concentrations (moles/L) Mol. Weight (grams/mol) Convert these mass concentrations to molar units (moles per liter) assuming density = 1.026 kg/L. Express the results using the appropriate M, mM, muM, nM etc. notation (for example, 1.35xl0^-3 M would be 1.35 mM, 1.35xl0^-2 would be 13.5 mM and 1.35xl0^-4 would be 135 mu M) What is the salinity of this sample?

Explanation / Answer

1 ppm = 1 mg solute per liter of solution= 0.001 g solute per liter of solution

To get molarity, molarity= moles/ Ltr of solution

So, 1 ppm can be converted into molarity terms as equal to (0.001 g solute per liter of solution/ Molar mass of solute).

For the above list of solutes, say for one example: Sodium

10700 ppm = 10.700 g of sodium per liter of solution/ 31 g/mol = 0.345 M of sodium.

You can do the calculations similarly for all other solutes.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.