3. The solubility of BaSO4 at 25C is 9.04 mg/L. Based on this, how many grams and moles of BaSO4 remained dissolved in the 100 mL of solution? What percent of your yield does this represent?
remained disolved in 100mL....has to be .904mg
I am not certain what your lab setup was.
To calculate the grams and moles of BaSO4 dissolved in the 100 mL solution, you'll need to use the solubility value provided.
First, convert 100 mL to liters:
100 mL = 100/1000 = 0.1 L
Next, multiply the solubility by the volume of the solution in liters:
Solubility (mg/L) * Volume of solution (L) = Mass of dissolved BaSO4 (mg)
9.04 mg/L * 0.1 L = 0.904 mg
Now, convert the mass of dissolved BaSO4 from milligrams to grams:
0.904 mg = 0.904/1000 = 0.000904 g
To convert grams to moles, you need to know the molar mass of BaSO4:
Ba has an atomic mass of 137.33 g/mol,
S has an atomic mass of 32.07 g/mol, and
O has an atomic mass of 16.00 g/mol.
So, the molar mass of BaSO4 is:
(1 * 137.33 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol) = 233.39 g/mol
Now, divide the mass of BaSO4 (in grams) by its molar mass to get moles:
0.000904 g / 233.39 g/mol ≈ 3.87 x 10^-6 mol
To calculate the percentage of your yield, you need to know the theoretical yield, or the maximum possible amount of BaSO4 that could have dissolved in the solution. Without additional information, we cannot determine the theoretical yield, so we cannot calculate the percent yield.