Answer :
Answer:
concentration of glucose = 78.75 mg/dL
Explanation:
The question essentially wants to test the ability to calculate the concentration of a patient's test result done on a spectrophotometer using the absorbance from Beer-Lambert's law, which states that when incident light passes through a medium, the absorbance is directly proportional to the concentration of the medium and inversely to the length of the light path.
Mathematically it is represented as
Absorbance (A) ∝ Concentration (C) (or length of path)
A ∝ C
A = kC
where "k" represents the factors that are kept constant.
As a result we can rewrite the formula as:
A₁ = C₁ - - - - - (1)
A₂ = C₂ - - - - -(2)
And dividing both equations:
[tex]\frac{A_1}{A_2} = \frac{C_1}{C_2}[/tex]
Next, let us define what is "standard" is; in analytical chemistry, a standard solution is one containing a precisely known concentration of the analyte in question, and it can be applied into the Beer-Lamberts law as follows:
[tex]\frac{A_T}{A_S} = \frac{C_T}{C_S} \\Where: \\A_S = Absorbance\ of\ test\\A_T = Absorbance\ of\ standard\\C_T = Concentration\ of\ test\\C_S = Concentration\ of\ standard[/tex]
[tex]Making\ C_T the\ subject\ of\ the\ equation\ by\ cross-multilication\\C_T \times A_S = A_T \times C_S\\C_T = \frac{A_T}{A_S} \times C_S\\where:\\ A_T = 0.252\\A_S = 0.640\\C_S = 200mg/dL\\\therefore C_T = \frac{0.252}{0.640} \times 200\\C_T = 78.75\ mg/dL[/tex]