Question

1. A standard solution of 100 ppm (100 mg / L) of an organic compound was analyzed and this has a maximum absorption band at 270 nm. A calibration curve was prepared from which the following data were obtained: Conc. (Ppm) 1 1,5 2 3 4 Absorbance 0.1282   0.1940 0.2615 0.3995 0.5422       
If 2 ml of an unknown solution of this compound are diluted to 50 ml, an absorbance of 0.425 is obtained.
a. Calculate the concentration of the sample in mg / L.
         
2. The organic compound 3-cyano-5-methylpyrroline (C7H6N2, MW: 118.14 g / mol) has an absorption band at 272 nm in ethanol solution. A solution containing 0.274 mg of this compound in 10 ml of ethanol gave an absorbance of 0.700 in a 1.00 cm cell.   Calculate: absorbability (5 points) the molar absorbability of the compound (5 points)                         
3.The% transmittance of a solution measured at a certain wavelength in a 1.00 cm cell is 63.5%. a.Calculate the% transmittance of this solution in an optical path cell of 2.0 mm.                 
4. A fragment of a silver coin weighing 0.1238 g was found. For its analysis it was dissolved in nitric acid and 0.0214 M KSCN was used as titrant, consuming of this 47.0 ml.   Calculate the percentage of silver content in the currency. It is considered that if a silver coin has a percentage less than 90% it is false. According to your results, the silver coin is original or false?

5. We want to analyze a sample of a physiological liquid using the complexometric titration method with EDTA as titrant. A sample of 100 ml of serum is taken and to this is added 2 drops of sodium hydroxide and immediately titrated with the EDTA 1.19 x 10 -3 M solution, which required 268 ml. Ca 2+ + Y 4- ---> CaY2-  
a.Determine the concentration of calcium in the sample as mg Ca / 100 ml serum              
6. Using the graph obtained in your spectrophotometric experiment for the determination of Iron, determine the concentration in ppm if a transmittance of 63% was obtained.

Answer 1

That is the absorbance vs concentration plot from the equation we got we know that

for an absorbance of 0.425

concentration is 3.168 ppm or 3.168 mg / L

that is the concentration of the 50 ml solution for the experiment

so apply dilution calculations

concentration 1 * volume 1 = concentration 2 * volume 2

Concentratio 1 is unknown

volume 1 is 2 ml

Concentration is 3.168

volume 2 is 50 ml (for the dilution) so

C1 * 2 = 3.168 * 50

C1 = 50*3.168 / 2 = 79.2 mg / L

This is the concentration of the unknown solution