(b) Quinine in a 3.925-g antimalarial tablet was dissolved in sufficient 0.10 M HCI to give 1.000 L. Dilution of a 20.00 mL aliquot to 100.00 mL yielded a solution that gave a reading of 415 at 347.5 nm. A second 20.00 mL aliquot was mixed with 10.0 mL of 50-ppm quinine solution before dilution to 100 mL. The fluorescence intensity of this solution was 503. Calculate the percentage of quinine in the tablet.
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- Caffeine, C8H10O2N4 H2O (MW = 212.2 g/mol) has been shown to have an average absorbance of 0.644 for a concentration of 1.783 mg per 100 mL at 272 nm. A sample of 3.658 g of a soluble coffee product was mixed with water to a volume of 500 mL and a 25 mL aliquot was transferred to a flask containing 25 mL of 0.1 M H2SO4. This was subjected to the prescribed clarification treatment and made up to 500 mL. A portion of this treated solution showed an absorbance of 0.666 at 272 nm. Assumbe b = 1.0 cm. Calculate the % caffeine (w/w) in the sample.3How do I find the theoretical molarity?
- The drug tolbutamide (MW 270.0g/mol), a treatment for type 2 diabetes, has a molar absorptivity of 703 /M/cm at 262nm. One tablet was dissolved in 250ml of water. A 10.0 ml aliquot of this solution was diluted to 100.00 ml in a volumetric flask. This diluted solution exhibited an absorbance of 0.275 at 262 nm in a 1.00 cm cell. What is the mass of tolbutamide in the tablet?g 2.png Average absorbance, .129 +0.128 +0.127)/3 = 0.128 Chrome Concentration of Females of Fe volume Q Molarity of Fe(NO₂)) = 0.2M Volume of Fe(NO₂)); = 5.00mL Moles of Fe³+= Molarity x Molarity x volume(L) Moles of Fe³* = 0.2M x 0.0054 = 10³mole 10-mole 500 x 10-3=0.002M [Fe³+] = [SCN¯] = X = ✪ [Fe³] = 0.002mol/L nknown concentrations to determine [FeSCN]at equilibrium in solution 8: Y = 4290X - 0.068 72% (0.127 + 0.068) 4290 0.127=4290X - 0.068 25ml Fe³+ x 0.002 mol/L-¹ 100mL 100% = 4.5 x 10-5mol/L 20ml SCN X 0.002 mol/L-¹ 100mL = 5.00 x 10^4 mol/L^-1 = 4.00 x 10 mol/L-1 W Flask 8 Flask 9 Flask 10 Flask 11 3.png 4.png Part 3: Q8 Flask 8 Flask 9 Flask 10 Flask 11 Part 3: 09 Flask 8 Flask 9 Flask 10 Flask 11 [Fe³+ (mol/L) 5.00 x 10-4 5.00 x 10-4 5.00 x 10-4 5.00 x 10-4 Average absorbance 0.127 0.166 0.222 69% 0.308 100% [FESCN¹ (mol/L) 4.5 x 10- 5.4 x 10 6.76 x 10-5 8.76 x 10- [SCN (mol/L) 4.00 x 10-4 4 x 10-4 7 x 10-4 1x 10-³ 3 7.png Place the results from calculations 9 and 10 into…Quinine in a 1.664 g antimalarial tablet was dissolved in sufficient 0.10 M HCl to give 500 mL of solution. A 15.00 mL aliquot was then diluted to 100.0 mL with the acid. The fluorescense intensity for the diluted sample at 347.5 nm provided a reading of 288 on an arbitrary scale. A standard 100 ppm quinine solution registered 180 when measured under conditions identical to those for the diluted sample. Calculate the mass in milligrams of quinine in the tablet.
- What is the concentration of a solution with an absorbance at 600 nm (A600) of 0.8 that has an extinction coefficient of 2.0 L/(mmol cm) (assume pathlength is 1 cm and provide your answer should in units of mM, but do NOT write the mM).Tonic water (20.0 mL) was added into each of two volumetric flasks (100 mL). The first flask was made up to volume with deionised water and produced an absorbance of 0.200. To the second flask was added an aliquot (20.0 mL) of a quinine standard solution (20.0 ppm) and after being made up to volume this produced an absorbance of 0.400. What is the concentration of quinine in the tonic water? a. 8.00 ppm b. 800 ppm С. 40.0 ppm d. 400 ppm е. 20.0 ppmA (0.002M) Cu2+ solution gave an absorbance of 0.50 at 560 nm in a 1.0 cm cell. The molar absobitivity of solution is equel:
- The following data are obtained from three standard soltions using UV-Vis spectroscopy: What is the colar concentration (M) of an unknown solution with an absorbance of 0.180 at 255 nm? The unknown solution contains the same analyte as the standard solution does. The molar mass of the analyte is 188.65 g/mol a) 4.82x10-5b)1.9x10-3c)3.60x10-4d)1.9x10-5e)1.8x 10-4A 131.7 mg sample of powdered Carbamazepine tablets was boiled with ethanol, filtered and then made up to 100 mL with ethanol. 5 mL of this solution was then diluted to 250 mL with ethanol and the absorbance at 285 nm found to be 0.486 in a 1cm path length cell.Calculate the content (% m/m) of Carbamazepine in the powder, taking A(1%, 1 cm) as 490.(State your answer to 1 decimal place and include the units)As part of an investigation on Pb in biological ecosystems, some grasses that grew along the road and were exposed to emissions from motor gasoline were collected. 6.2501 g of grass were calcined to destroy all organic matter, the inorganic residue was properly treated and finally made up to 100 mL. An aliquot of 50 mL of this solution was read in an atomic absorption spectrophotometer registering an absorbance of 0.1250; then 0.10 mL of a standard solution of 55 µg of Pb / mL were added to the other 50 mL, with this fortified sample, another absorbance measurement was made, which was 0.1798. Calculate the µg of Pb / g of grass