How many grams of beryllium fluoride are needed to prepare 250.0 mL of 0.224 M F−?

Question

You wish to prepare 250.0 mL of a 350.0 ppm w/v fluoride (MW = 18.9984 g/mol) solution. How many grams of beryllium fluoride ( BeF2 , MW = 47.01 g/mol) are needed to prepare this solution? Assume the final solution has a density of 1.00 g/mL. How many grams of beryllium fluoride are needed to prepare 250.0 mL of 0.224 M F−?

Answer 1

350.0 ppm F^- is the same as 350.0 mg F^- per liter

250.0 ml = 0.2500 L

350.0 mg F^- / L x 0.2500 L = 87.50 mg F^- needed

Molar mass BeF₂ = 47.01 g / mol

% F by mass = 2 x 18.9984 g / 47.01 g (x100%) = 37.9968 g / 47.01 g (x100%) = 80.83% F

80.83% x = 87.50 mg

x = 108.6 mg of BeF₂ needed

108.6 mg BeF₂ x 1 g / 1000 mg = 0.1086 g BeF₂ needed

To prepare 250.0 ml of 0.224 M F^-:

BeF₂ ==> Be²⁺ + 2F^- (note, 2 F^- per 1 BeF₂)

0.224 mol F^-/ L x 1 mol BeF₂ / 2 mol F^- x 0.250 L x 47.01 g / mol = 1.32 g BeF₂