The technician began his analysis by preparing a stock solution of fluorescein. To prepare the fluorescein stock solution, 0.01140 g of disodium fluorescein salt (MM = 376.28 g/mol) was weighed on the microanalytical balance. The salt was dissolved in 50 mL of 0.1 M NaOH and then analytically transferred to a 100.00 mL volumetric flask. The flask was diluted to the mark with 0.1 M NaOH. Calculate the concentration of fluorescein (MM = 332.21 g/mol) stock solution in ppm.
The technician began his analysis by preparing a stock solution of fluorescein. To prepare the fluorescein stock solution, 0.01140 g of disodium fluorescein salt (MM = 376.28 g/mol) was weighed on the microanalytical balance. The salt was dissolved in 50 mL of 0.1 M NaOH and then analytically transferred to a 100.00 mL volumetric flask. The flask was diluted to the mark with 0.1 M NaOH. Calculate the concentration of fluorescein (MM = 332.21 g/mol) stock solution in ppm.
The Correct Answer and Explanation is:
Answer:
The concentration of fluorescein in the final stock solution is 100.6 ppm.
Step-by-Step Calculation and Explanation (≈300 words):
To calculate the concentration in ppm (parts per million) of fluorescein in the stock solution, we follow these steps:
1. Determine moles of disodium fluorescein salt used:
Given:
- Mass of disodium fluorescein salt = 0.01140 g
- Molar mass (MM) of disodium fluorescein salt = 376.28 g/mol
Moles of disodium salt=0.01140 g376.28 g/mol=3.03×10−5 mol\text{Moles of disodium salt} = \frac{0.01140 \text{ g}}{376.28 \text{ g/mol}} = 3.03 \times 10^{-5} \text{ mol}
2. Convert to moles of fluorescein:
Disodium fluorescein salt dissociates to form fluorescein (C₂₀H₁₀O₅), so the mole ratio is 1:1.
Thus, moles of fluorescein = 3.03 × 10⁻⁵ mol
3. Calculate mass of fluorescein:
Given:
- MM of fluorescein = 332.21 g/mol
Mass of fluorescein=3.03×10−5 mol×332.21 g/mol=0.01006 g\text{Mass of fluorescein} = 3.03 \times 10^{-5} \text{ mol} \times 332.21 \text{ g/mol} = 0.01006 \text{ g}
4. Calculate concentration in ppm:
ppm is defined as mg of solute per liter (mg/L) of solution.
- Final volume of stock solution = 100.00 mL = 0.10000 L
- Convert 0.01006 g to mg:
0.01006 g=10.06 mg0.01006 \text{ g} = 10.06 \text{ mg}
Now: ppm=10.06 mg0.10000 L=100.6 ppm\text{ppm} = \frac{10.06 \text{ mg}}{0.10000 \text{ L}} = 100.6 \text{ ppm}
Summary:
- The disodium fluorescein salt was converted to pure fluorescein after dissolving.
- Moles were calculated using the salt’s molar mass.
- Fluorescein mass was then calculated using its own molar mass.
- The final step converted that mass to ppm based on the 100 mL total solution volume.
This approach properly accounts for the dissociation of the salt and uses the molar mass of fluorescein (not the salt) in the final ppm concentration calculation.
