You wish to prepare 100.0 ml of a 525.0 ppm w/v fluoride (mw = 18.9984 g/mol) solution. how many grams of beryllium fluoride ( bef 2 , mw = 47.01 g/mol) are needed to prepare this solution? assume the final solution has a density of 1.00 g/ml.

0.06495 g of are needed to prepare this solution.

Explanation:

1 ppm = 1 mg/L

525.0 ppm W/V fluoride solution means 1 L of fluoride solution contains 525.0 mg of .

So, 100.0 mL (0.1000 L) solution contains of or 52.50 mg of .

Molar mass of = 19.00 g

2 moles of are present in 1 mol of

So, 38.00 g of are present in 47.01 of

Hence 0.05250 g of are present in of or 0.06495 g of

where is the following

explanation: