Question

The absorbance (????)(A) of a solution is defined as ????=log10(????0????) A=log10⁡(I0I) where ????0I0 is the incident‑light intensity and ????I is the transmitted‑light intensity. Absorbance is also defined as ????=????c???? A=ϵcl where ????ϵ is the molar absorption coefficient (extinction coefficient) in units of M−1cm−1,M−1cm−1, cc is the molar concentration, and ????l is path length in centimeters. Daniella prepares a 1 mg/ml1 mg/ml myoglobin solution. The molecular weight of myoglobin is 17.8 kDa.17.8 kDa. Given that the ????ϵ of myoglobin is 15,000 M−1cm−1,15,000 M−1cm−1, calculate the absorbance of the myoglobin solution across a 1 cm1 cm path. Calculate your answer to two decimal places.

Answer 1

The absorbance of the myoglobin solution across a 1 cm path is 0.84.

Step-by-step explanation:

Beer-Lambert's law :

Formula used :

`A=\epsilon * c* l`

`A=\log (I_o)/(I)`

`\log (I_o)/(I)=\epsilon * c* l`

where,

A = absorbance of solution

c = concentration of solution

`\epsilon` = Molar absorption coefficient

l = path length

`I_o` = incident light

`I` = transmitted light

Given :

l = 1 cm, c = 1 mg/mL ,
`\epsilon = 15,000 M^(-1)cm^(-1)`

Molar mass of myoglobin = 17.8 kDa = 17.8 kg/mol=17800 g/mol

(1 Da = 1 g/mol)

c = 1 mg /mL =
`{1mg /mL}{\text{Molar mass of myoglobin}}`

`c = (1 mg/mL)/( 17800 g/mol) = 5.6179* 10^(-5) mol/L`

1 mg = 0.001 g, 1 mL = 0.001 L

`A= 15,000 M^(-1)cm^(-1)* 5.6179* 10^(-5) mol/L* 1 cm`

`A=0.8426 \approx 0.84`

The absorbance of the myoglobin solution across a 1 cm path is 0.84.