Question

1 31% all VZW Wi-Fi 20:30 Exit 55:33:15 3. Calculate parts per million Br in a sample of seawater that is 0.865 mM Br witha density of 1.025 g/mL 69.1 ppm 70.8 ppm 67.4 ppm 94.7 ppm 79.9 ppm Submit

Answer 1

The concentration of the Br sample = 0.865 mM = 0.865*10^-3 mol/L.

Hence, 1 L of the Br ^- solution has 0.865*10^-3 mol Br.

Atomic mass of Br^- = 79.904 g/mol

Therefore, mass of Br^- in the sample = (0.865*10^-3 mol)*(79.904 g/mol) = 0.0691 g = (0.0691 g)*(1000 mg/1 g) = 69.1 mg.

Density of solution = 1.025 g/mL; therefore, mass of 1 L = 1000 mL of seawater = (1.025 g/mL)*(1000 mL) = 1025 g = (1025 g)*(1 kg/1000 g) = 1.025 kg.

Concentration of Br- in ppm = (milligrams of Br)/(kilograms of seawater) = (69.1 mg)/(1.025 kg) = 67.4 mg/kg = 67.4 ppm (ans)

Ans: (E) 67.4 ppm