Question

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.

Answer 1

Answer:

0.06495 g of $BeF_{2}$ 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 $F^{-}$.

So, 100.0 mL (0.1000 L) solution contains $(525.0\times 0.1000)mg$ of $F^{-}$ or 52.50 mg of $F^{-}$.

Molar mass of $F^{-}$ = 19.00 g

2 moles of $F^{-}$ are present in 1 mol of $BeF_{2}$

So, 38.00 g of $F^{-}$ are present in 47.01 of $BeF_{2}$

Hence 0.05250 g of $F^{-}$ are present in $\frac{47.01\times 0.05250}{38.00}$ of $BeF_{2}$ or 0.06495 g of $BeF_{2}$