What volume of 6.83% (wt./vol.) NaOCl solution is required to oxidize 223 mg of 9-fluorenol to 9-fluorenone?
Please help me set this up!
To determine the volume of NaOCl solution required, we can use the following setup:
Step 1: Calculate the moles of 9-fluorenol.
First, convert the mass of 9-fluorenol from milligrams (mg) to grams (g):
223 mg = 0.223 g
Next, determine the molar mass of 9-fluorenol by adding up the atomic masses of its constituent elements from the periodic table. The molar mass of 9-fluorenol is 180.23 g/mol.
Now, use the formula:
moles = mass / molar mass
moles of 9-fluorenol = 0.223 g / 180.23 g/mol
Step 2: Determine the stoichiometry between 9-fluorenol and NaOCl.
The balanced chemical equation for the oxidation of 9-fluorenol to 9-fluorenone using NaOCl is:
3 C13H10OH + 5 NaOCl -> 3 C13H8O + 5 NaOH
From the equation, we can see that the stoichiometric ratio between 9-fluorenol and NaOCl is 3:5.
Step 3: Calculate the moles of NaOCl required.
Using the stoichiometric ratio, the moles of NaOCl required can be determined:
moles of NaOCl = (moles of 9-fluorenol) * (5/3)
Step 4: Calculate the volume of NaOCl solution.
The concentration of the NaOCl solution is given as 6.83% (wt./vol.), which means there are 6.83 grams of NaOCl in 100 mL of solution.
Since we know the moles of NaOCl required, the volume of NaOCl solution can be determined using the formula:
volume (in mL) = (moles of NaOCl) * (molar mass of NaOCl) / (concentration in g/mL)
Finally, plug in the values and calculate the volume of NaOCl solution needed.