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.

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-

And a) oxygen i believe