Sea water, which can be considered as a 6 molar (6 M) solution of NaCl , has a density of 2 g mL⁻¹. The concentration of dissolved oxygen (O₂) in sea water is 5.8 ppm . Then the concentration of dissolved oxygen (O₂) in sea water, is x × 10⁻⁴ m.
x= ___________. (Nearest integer)
Given: Molar mass of NaCl is 58.5 g mol⁻¹
Molar mass of O₂ is 32 g mol⁻¹

Solution

Chemistry ma’am · 4 steps

Take exactly 1 L of sea water, find the grams of water in it, then turn 5.8 ppm into moles of O₂.

1. Mass of 1 L. With density 2 g mL⁻¹, 1 L of sea water weighs 1000 × 2 = 2000 g. It holds 6 mol NaCl, which is 6 × 58.5 = 351 g.
2. Mass of solvent. Molality needs the mass of the solvent, not of the solution. Water = 2000 - 351 = 1649 g = 1.649 kg. The tiny mass of dissolved O₂ does not change this.
3. Moles of oxygen. 5.8 ppm means 5.8 g of O₂ in 10⁶ g of solution. In 2000 g that is (5.8 × 2000)/10⁶ = 0.0116 g, which is 0.0116/32 = 3.625 × 10⁻⁴ mol.

   m = (3.625 × 10⁻⁴ mol)/(1.649 kg) = 2.2 × 10⁻⁴ mol kg⁻¹

4. So. The molality of dissolved O₂ is about 2.2 × 10⁻⁴ m, so x rounds to 2.

The idea to keep. For molality from molarity and density, fix 1 L of solution, find its mass, subtract the solute mass, and divide moles by kilograms of solvent. ppm is mass of solute per 10⁶ g of solution.

Answer from the NTA final answer key. Solution by PrepWiser.