Gockululian Mass of Fluorescein Molar mass of Fluorescein sodium Volume 9590 ethanol Concentration Methylene Black solution Concentration of fluorescein stock solution ppm mol/L Preparation of standard solutions Fluorescein solution Fluorescein solution Fluorescein solution Fluorescein solution Fluorescein solution Fluorescein solution 150 06 ppmL ppI pu ppm ppm ppm The absorption and emission spectra of fluorescein. The calibration curve Fluorescence MX Solution concentration Fluorescence intensity Calculation of the ppm concentration for unknown mol
Gockululian Mass of Fluorescein Molar mass of Fluorescein sodium Volume 9590 ethanol Concentration Methylene Black solution Concentration of fluorescein stock solution ppm mol/L Preparation of standard solutions Fluorescein solution Fluorescein solution Fluorescein solution Fluorescein solution Fluorescein solution Fluorescein solution 150 06 ppmL ppI pu ppm ppm ppm The absorption and emission spectra of fluorescein. The calibration curve Fluorescence MX Solution concentration Fluorescence intensity Calculation of the ppm concentration for unknown mol
The Correct Answer and Explanation is:
To provide a meaningful answer and explanation to your query, I will assume you are conducting a fluorescence spectroscopy experiment involving fluorescein and possibly methylene blue, where you are:
- Preparing standard solutions of fluorescein,
- Measuring their fluorescence intensity,
- Constructing a calibration curve, and
- Using this to calculate the concentration of an unknown fluorescein solution in ppm and mol/L.
Correct Answer with Example Calculation
Given/Assumed Data:
- Mass of fluorescein sodium used: 150 mg = 0.150 g
- Molar mass of fluorescein sodium: 376.27 g/mol
- Volume of ethanol used to dissolve fluorescein: 95.90 mL = 0.0959 L
- Concentration of Methylene Blue solution: Not relevant unless used as a comparison or quencher
- Goal: Calculate stock solution concentration in ppm and mol/L
1. Moles of Fluorescein Sodium:
Moles=MassMolar Mass=0.150376.27=3.987×10−4 mol\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.150}{376.27} = 3.987 \times 10^{-4} \text{ mol}
2. Concentration in mol/L:
Concentration (mol/L)=3.987×10−40.0959=4.16×10−3 mol/L\text{Concentration (mol/L)} = \frac{3.987 \times 10^{-4}}{0.0959} = 4.16 \times 10^{-3} \text{ mol/L}
3. Convert to ppm (mg/L):
- Since 1 mol of fluorescein sodium = 376.27 g = 376,270 mg:
Concentration in ppm=4.16×10−3 mol/L×376270 mg/mol=1565 ppm\text{Concentration in ppm} = 4.16 \times 10^{-3} \text{ mol/L} \times 376270 \text{ mg/mol} = 1565 \text{ ppm}
Explanation
In fluorescence spectroscopy, accurate preparation and quantification of the fluorophore (here, fluorescein sodium) are essential for producing reliable results. Fluorescein is a widely used fluorescent dye due to its strong fluorescence emission under UV light, and it is often used in tracing, biomedical imaging, and analytical chemistry.
To prepare a stock solution, a known mass of fluorescein sodium—150 mg in this case—is dissolved in a solvent (95.90 mL of ethanol). The molar mass of fluorescein sodium (376.27 g/mol) allows us to convert this mass into moles, yielding 3.987 × 10⁻⁴ mol. Dividing this by the solution volume gives a molarity of 4.16 × 10⁻³ mol/L.
For practical comparisons and calibration, we often express concentration in parts per million (ppm), especially when dealing with dilute solutions. Using the molarity and the molar mass, we calculate a concentration of approximately 1565 ppm. This solution can then be used to prepare a dilution series (e.g., 150, 60, 30, 15 ppm, etc.) to build a calibration curve by plotting fluorescence intensity versus known concentration.
This curve is crucial: once the fluorescence intensity of an unknown sample is measured, its concentration can be determined using the linear relationship from the calibration plot.
This approach ensures accurate quantification of fluorescein in unknown samples and highlights the utility of fluorescence as a sensitive and selective analytical technique.
