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

Answer:

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

Explanation:

Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times c\times l$

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

$c = \frac{1 mg/mL}{ 17800 g/mol} = 5.6179\times 10^{-5} mol/L$

1 mg = 0.001 g, 1 mL = 0.001 L

$A= 15,000 M^{-1}cm^{-1}\times 5.6179\times 10^{-5} mol/L\times 1 cm$

$A=0.8426 \approx 0.84$

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