Question

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)?

Answer 1

Answer:

The answer is "$4.875 \ \frac{mol}{l}$".

Explanation:

The molar absorptivity value is=  12.3    $\frac {L}{mol \times cm}$

path length of cell = 1  $\ cm$

Absorbance = 1.2

Using the beer's lambert law:

$A = \varepsilon \ cl\ \ \ \ \ \ \ _{where} \\\\c= \frac{A}{ \varepsilon l }$

  $= \frac{1.2}{12.3 \times 1}\\\\= 0.0975 \ \frac{mol}{L}$

Claculating the concentration of the solution after dilution: $(m_2)= 0.0975 \ \ \frac{mol}{L}$

find out the dilution value before concentration :

The volume taken by the dilution: $(V_1) = 0.2\ ml$

The final volume after dilution:

$(V_2) = 9.8+0.2$

       $= 10 \ ml$

Formula:

$\bold{M_1 \times V_1 = M_2 \times V_2} \\\\ M_1= \frac{M_2 \times V_2 }{V_1}$

     $= \frac{ 0.0975 \times 10 }{0.2}\\\\= \frac{ 0.975 }{0.2}\\\\ = 4.875 \ \frac{mol}{l}$