Question

Express the concentration of a 0.0610 m aqueous solution of fluoride, f–, in mass percentage and in parts per million. assume the density of the solution is 1.00 g/ml.

Answer 1

The solution concentration of 0.0610 m aqueous fluoride is 0.1159% by mass and 1159 ppm. These values are calculated using the molarity and molar mass of fluoride and assuming a solution density of 1.00 g/mL.

Step-by-step explanation:

To express the concentration of a 0.0610 m aqueous solution of fluoride (F−) in mass percentage and parts per million (ppm), given a density of 1.00 g/mL, we need to do some calculations based on the definitions of these concentration units.

First, to find the mass percentage, we use the molarity of the solution (moles of solute per liter of solution) and the molar mass of fluoride (F−, approximately 19.00 g/mol). Since the solution has a density of 1 g/mL, we assume that 1 liter of the solution weighs 1000 g.

The mass of fluoride in 1 L of 0.0610 M solution is:

0.0610 moles/L × 19.00 g/mole = 1.159 g of F−

To express this as a percentage, we divide the mass of fluoride by the total mass of the solution and multiply by 100:

(1.159 g / 1000 g) × 100 = 0.1159%

Now, to convert this to parts per million, we use the fact that 1 ppm equals 1 mg of solute per kg of solution. Considering that 1.159 g is 1159 mg, we have:

1159 mg of F− per 1000 g of solution = 1159 ppm

Therefore, the concentration of the fluoride solution is 0.1159% by mass and 1159 ppm.

Answer 2

Assume you have 1 L of solution.
Moles F- = M F- x L F- = (0.0610)(1) = 0.0610 moles F-
0.0610 moles F- x(19.0 g F-/1mole F-) = 1.159 g F- in 1 L of solution
1 L solution x (1000 mL / 1 L) x (1.00 g / mL) = 1000 g of solution
mass % F- = (g F- / g solution) x 100 = (1.159 / 1000) x 100
= 0.1159%
parts per million F- = mg F- /L = 1159 / 1 = 1159 ppm F-