Question

Express the concentration of a 0.0390 M aqueous solution of fluoride, F − , in mass percentage and in parts per million (ppm). Assume the density of the solution is 1.00 g/mL.

Answer 1

Answer:

Mass percentage → 0.074 %

[F⁻] = 741 ppm

Explanation:

Aqueous solution of flouride → [F⁻] = 0.0390 M

It means that in 1L of solution, we have 0.0390 moles of F⁻

We need the mass of solution and the mass of 0.0390 moles of F⁻

Mass of solution can be determined by density:

1g/mL = Mass of solution / 1000 mL

Note: 1L = 1000mL

Mass of solution: 1000 g

Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g

Mass percentage → (Mass of solute / Mass of solution) . 100

(0.741 g / 1000 g) . 100 = 0.074 %

Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)

0.741 g . 1000 mg/1g = 741 mg

1000 g . 1 kg/1000 g = 1kg

741 mg/1kg = 741 ppm