A sample of unknown concentration is diluted by placing 0.2 mL into 9.8 mL of buffer and then 3 mL is placed into a standard 1x1 cm square cuvette. It is measured at 350 nm and the Abs is 0.3. The molar absorptivity is 12.3 L/mol*cm. What is the concentration of the undiluted sample (mol/L to 2 decimal places)?

Respuesta :

Answer:

The answer is "[tex]4.875 \ \frac{mol}{l}\\[/tex]".

Explanation:

The molar absorptivity value is=  12.3    [tex]\frac {L}{mol \times cm}[/tex]

path length of cell = 1  [tex]\ cm[/tex]

Absorbance = 1.2

Using the beer's lambert law:

[tex]A = \varepsilon \ cl\ \ \ \ \ \ \ _{where} \\\\c= \frac{A}{ \varepsilon l } \\\\[/tex]

  [tex]= \frac{1.2}{12.3 \times 1}\\\\= 0.0975 \ \frac{mol}{L}\\[/tex]

Claculating the concentration of the solution after dilution: [tex](m_2)= 0.0975 \ \ \frac{mol}{L}[/tex]

find out the dilution value before concentration :

The volume taken by the dilution: [tex](V_1) = 0.2\ ml[/tex]

The final volume after dilution:

[tex](V_2) = 9.8+0.2\\[/tex]

       [tex]= 10 \ ml[/tex]

Formula:

[tex]\bold{M_1 \times V_1 = M_2 \times V_2} \\\\ M_1= \frac{M_2 \times V_2 }{V_1}\\\\[/tex]

     [tex]= \frac{ 0.0975 \times 10 }{0.2}\\\\= \frac{ 0.975 }{0.2}\\\\ = 4.875 \ \frac{mol}{l}\\[/tex]