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

The answer is "
`4.875 \ (mol)/(l)\\`".

Step-by-step explanation:

The molar absorptivity value is= 12.3
`\frac {L}{mol * cm}`

path length of cell = 1
`\ cm`

Absorbance = 1.2

Using the beer's lambert law:

`A = \varepsilon \ cl\ \ \ \ \ \ \ _(where) \\\\c= (A)/( \varepsilon l ) \\\\`

`= (1.2)/(12.3 * 1)\\\\= 0.0975 \ (mol)/(L)\\`

Claculating the concentration of the solution after dilution:
`(m_2)= 0.0975 \ \ (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 * V_1 = M_2 * V_2} \\\\ M_1= (M_2 * V_2 )/(V_1)\\\\`

`= ( 0.0975 * 10 )/(0.2)\\\\= ( 0.975 )/(0.2)\\\\ = 4.875 \ (mol)/(l)\\`