5. It might make it easier to understand the optimum range for absorbance measur

Question

5. It might make it easier to understand the optimum range for absorbance measurements using the colorimeter (0.05 A 1.0) if you compare it to your eyes, your vision Sensor The analogy: If you stare directly into the sun on a bright day, you are blinded", your eye is "oversaturated" with photons. You would be unable to determine the color of an object. When absorbance is very low, most of the light is going through the sample to the colorimeter's detector. The detector is "oversaturated" with photons, so its accuracy suffers. Your eyesight is poor in the dark, because of the low number of photons reaching your eye.

Explanation / Answer

The regression equation is given as

y = 0.0103*C – 0.025 (C has units of mg/L).

First find the value of C from solution #3 using the equation above. Put y = 0.73 and obtain

0.73 = 0.0103*C – 0.025

===> 0.0103*C = 0.755

===> C = 73.30

The concentration of solution #3 is 73.30 mg/L (ans).

Next find the concentration of solution #1. Note that the absorbance was offscale (meaning that the concentration was higher than the limit of detection). So, we will use the dilution equation. Check that 1.00 mL of solution #1 was taken and diluted to 25.00 mL to obtain solution #3.

Use the dilution equation to find the concentration of solution #1 (call this C1).

(1.00 mL)*C1 = (25.00 mL)*(73.30 mg/L)

===> C1 = (25.00*73.30)/(1.00) mg/L = 1832.5 mg/L.

The concentration of solution #1 is 1832.5 mg/L.

Finally find the concentration of the stock solution (call this C2).

(10.00 mL)*C2 = (25.00 mL)*(1832.5 mg/L)

===> C2 = (25.00*1832.5)/(10.00) mg/L = 4581.25 mg/L = (4581.25 mg/L)*(1 g/1000 mg) = 4.58125 g/L.

The concentration of the stock solution is 4.58125 g/L 4.58 g/L (ans).

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