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Chrome File Edit View History Bookmarks People Window Help 39% D Sun 6:53 PM Brianna Jimenez E CChegg.com x D chem Team: Converting be x ZConvert mmol to mole co x Brianna VA Experiment 5: Chem 7 Part X miami marlins standings X C fi https:// 13923199 3. 4.28/33.34 points I Previous Answers My Notes Ask Your Teacher E BUN Solution Absorbance You measure the %T values for a series of Six BUN Standard solutions. You need to convert the values to absorbance to determine its relationship to the solution concentration. Solution [BUN] (PPM) [BUNJ (M) %T Absorbance 7.59 1.12 550 275 27.5 .561 52.2 .282 138 72.3 .141 69.0 5 85.1 .07 34.5 6 92.2 .035 17.30 Analysis of an Unknown Sample You process your patient's BUN sample and determine its Transmittance to be 88.0. What is the concentration of BUN in your patient's sample? (Note: The patient is a male in his mid-forties mg/dL Is this reading high, low or normal? high low norma Submit Answer Save Progress Practice Another Version

Explanation / Answer

The relation between absorbance(A) & % transmitance (% T) is A = log10 (100 / %T)

Also A=ebc

Where

A = absorbance

e = molar absorbtivity with units of L mol-1 cm-1

b = path length of the sample = path length of the cuvette in which the sample is contained

c = concentration of the compound in solution, expressed in mol L-1

For the first values in the given tabular form , A = ecl

A = 1.12 , c = 550 ppm               Since 1M = 35500 ppm ---> 1 ppm = (1/35500)M

                   = 550 x(1/35500) M

                   = 0.0155 M

l = length of the cuvette = 1cm

Plug the values we get e = A / (cl)

                                    = 72.3 L mol-1 cm-1

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Calculation of Absorbance at 88.0 % T :

A = log10 (100 / %T)

A = log10 (100 /88.0)  

   = 0.0555

So c = A / (el)

       = 0.0555 / (72.3x1)

      = 7.68x10-4 M

      = 7.68x10-4 x35500 ppm

     = 27.2 ppm

     = 27.2 x 0.1 mg/dL                since 1mg/dL = 10 ppm

     = 2.72 mg/dL

Therefore the concentration of BUN is 2.72 mg/dL

From the tabular form & the obtained value of A & c the value is normal

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