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

To prepare a 525.0 ppm w/v fluoride solution with a volume of 100.0 mL, you would need 52.5 grams of beryllium fluoride (BeF2).

Step-by-step explanation:

To prepare a 525.0 ppm w/v fluoride solution with a volume of 100.0 mL, we need to determine the mass of beryllium fluoride (BeF2) needed. The formula for a ppm w/v solution is:

ppm w/v = (mass of solute in grams / volume of solution in mL) x 106

Rearranging the formula, we can solve for the mass of solute:

mass of solute in grams = ppm w/v x volume of solution in mL / 106

Substituting the given values, we have:

mass of solute in grams = 525.0 x 100.0 / 106

mass of solute in grams = 52.5 g

Therefore, we need 52.5 grams of beryllium fluoride to prepare the solution.

Answer 2

0.06495 g of
`BeF_(2)` are needed to prepare this solution.

Step-by-step 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* 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
`(47.01* 0.05250)/(38.00)` of
`BeF_(2)` or 0.06495 g of
`BeF_(2)`