To solve this question, we'll follow these steps:
1. Determine the volume of one drop in milliliters (mL).
2. Calculate the number of moles of bromide ions in one drop.
3. Convert the number of moles to grams using the molar mass of bromide ions.
Step 1: Determine the Volume of One Drop
Given that it takes 145 drops to reach the 5.0 mL mark on a graduated cylinder, the volume of one drop (V_drop) in milliliters is given by dividing the total volume by the number of drops:
[tex]\[ V_{\text{drop}} = \frac{5.0\, \text{mL}}{145\, \text{drops}} \][/tex]
Step 2: Calculate the Number of Moles of Bromide Ions in One Drop
We are given that the molarity of the iron III bromide solution is 0.1 M. This means there are 0.1 moles of iron III bromide in 1 liter (1000 mL) of the solution. Since iron III bromide dissociates into one iron ion and three bromide ions, there will also be 0.1 * 3 = 0.3 moles of bromide ions in 1 liter of the solution.
To find the number of moles of bromide ions in one drop (n_drop), we use the relation:
[tex]\[ n_{\text{drop}} = V_{\text{drop}} \times \text{Molarity} \][/tex]
But since the molarity provided is 0.1 M for iron III bromide, the molarity for bromide ions will be 0.3 M.
Remember to convert milliliters to liters in this calculation (1000 mL = 1 L):
[tex]\[ n_{\text{drop}} = \left( \frac{V_{\text{drop}}}{1000} \right) \times 0.3 \][/tex]
Step 3: Convert Number of Moles to Grams
Lastly, we need to use the molar mass of the bromide ion (Br-) to convert moles to grams. The molar mass of a bromide ion is approximately 79.904 grams per mole.
The mass of bromide ions in one drop (mass_drop) is then calculated by multiplying the number of moles in one drop by the molar mass of bromide:
[tex]\[ \text{mass}_{\text{drop}} = n_{\text{drop}} \times \text{Molar mass of bromide} \][/tex]
Let's put these steps into numbers:
Step 1:
[tex]\[ V_{\text{drop}} = \frac{5.0\, \text{mL}}{145\, \text{drops}} \][/tex]
[tex]\[ V_{\text{drop}} \approx 0.03448\, \text{mL} \][/tex]
Step 2:
[tex]\[ n_{\text{drop}} = \left( \frac{0.03448\, \text{mL}}{1000} \right) \times 0.3 \][/tex]
[tex]\[ n_{\text{drop}} \approx \left( 0.03448 \times 10^{-3} \right) \times 0.3 \][/tex]
[tex]\[ n_{\text{drop}} \approx 1.0344 \times 10^{-5}\, \text{mol} \][/tex]
Step 3:
[tex]\[ \text{mass}_{\text{drop}} = 1.0344 \times 10^{-5}\, \text{mol} \times 79.904\, \text{g/mol} \][/tex]
[tex]\[ \text{mass}_{\text{drop}} \approx 8.2714 \times 10^{-4}\, \text{g} \][/tex]
Thus, there are approximately [tex]\( 8.2714 \times 10^{-4} \)[/tex] grams, or 0.00082714 grams, of bromide ions in one drop of the 0.1 M iron III bromide solution.
