Final answer:
Approximately 6.03 grams of boric acid are required to make 4 fl oz of the solution when 51 g of boric acid is used to make 1000 mL.
Step-by-step explanation:
To determine how many grams of boric acid are required to make 4 fl oz of a boric acid solution where 51 g of boric acid is used to make 1000 mL, we need to perform a unit conversion and proportion calculation.
First, we convert 4 fl oz to mL, knowing that 1 fl oz is approximately 29.5735 mL. Thus, 4 fl oz is about 118.294 mL. Next, we use the given concentration of the original solution (51 g per 1000 mL) to find out how much boric acid will be needed for our new volume.
The calculation is as follows:
- (51 g boric acid / 1000 mL solution) x 118.294 mL = grams of boric acid needed
Performing this calculation:
- 51 g/1000 mL * 118.294 mL ≈ 6.03 g
Therefore, approximately 6.03 grams of boric acid are required to make 4 fl oz of the solution.