Questions: (a) Calculate the molar concentration of glucose in the stock solutio

Question

Questions:

(a) Calculate the molar concentration of glucose in the stock solution.

(b) For each standard solution and the unknown, convert the percent transmittance to absorbance, using this equation: A=2.000 - log(%T). Enter these absorbences in the table.

(c) Calculate the molar concentration of glucose complex present in each solution. Enter the concentrations in the table.

(d) Prepare a Beer's law plot, using the data obtained in (b) and (c).

(e) Using your Beer's law plot, determine the molar concentration of glucose in the unknown sample

The determination of glucose concentration in blood serum is often based on the formation of a blue-green complex of glucose and o-toluidine in glacial acetic acid. This reaction is shown at the bottom of this page. The analytical wavelength for the complex is 635 nm. A set of standard solutions was prepared by taking a known volume of a stock glucose solution, containing 1.000 g glucose dissolved in 100 mL of distilled water, and diluting to 100 mL with distilled water. These solutions and the blood samples were treated with the o-toluidine reagent, and the percent transmittance of each sample was read against a reference solution. The data obtained are

Explanation / Answer

(a) Molar concentration of glucose in stock solution;

MW of glucose = 180.1559 g/mol

So 1 gm glucose contains 1/180.1559 moles of glucose

This much moles are present in 100 mL = 0.1 L

So molar concentration = (1/180.1559) / 0.1 mol/L or M = 0.0555 M

(b) Absorbance, A = 2 - log(%T) :

2 - log(76.91) = 2 - 1.886 = 0.114

Similarly,

2 - log(59.16) = 2 - 1.772 = 0.228

2 - log(35.08) = 2 - 1.545 = 0.455

2 - log(27.10) = 2 - 1.433 = 0.567

2 - log(20.65) = 2 - 1.315 = 0.685

2 - log(46.17) = 2 - 1.664 = 0.336

(c) Molar concentration of glucose complex:

5 mL x 0.0555 M / (100 + 5) mL = 0.0026 M

Similarly,

10 mL x 0.0555 M / (100 + 10) mL = 0.0050 M

20 mL x 0.0555 M / (100 + 20 mL = 0.0093 M

25 mL x 0.0555 M / (100 + 25) mL = 0.0111 M

30 mL x 0.0555 M / (100 + 30) mL = 0.0128 M

(d) Plot the concentrations from 'c' as x-axis and aborbance from 'b' as y-axis. And it will be a straight line. Do straight line curve fitting and you'll get the equation of the straight line.

I don't know how to attach pictures, but I did the excel plot for you (pretty simple).

The equation of the plot: y = 55.405 x - 0.0423 with R2 = 0.9961

(e) Find out the x-value corresponding to absorbance of unknown (0.336) and that will be your answer.

When y = 0.336,

0.336 = 55.405 x - 0.0423

or 55.405 x = 0.3783

or x = 0.0068 M

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