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The absorbance of a chlorophyll a standard solution in ethanol (density = 0.785 g/mL) is measured in

a 1.3 cm cuvette or path length. The absorbance is 0.765 and the molar absorptivity coefficient is 86,300 M–1cm–1.

The molecular weight for chlorophyll a is 893.49 g/mol.

What is the concentration of the solution in ppm?

Verified Answer :

syed514

Let A= ebc 
     0.765 = 86,3001.3c 

Solve for c = approximately 7E-6 Molar
                  = mols/L soln. 
g = 7E-6893.49
   = about 0.006 g chlorophyll/L soln. 

1000 x 0.785 = 785 g ethanol. 
Conc. = about 0.006g chlorophyll/785 g soln. 

Change that to ppm. by using formula:
(0.006/785)1E6

Answer:

7.76 is the concentration of the solution in ppm.

Explanation:

Using Beer-Lambert's law :

Formula used :

\(A=\epsilon \times C\times l\)

where,

A = absorbance of solution = 0.765

C = concentration of solution = ?

l = path length = 1.3 cm

\(\epsilon\) = molar absorptivity coefficient = \(86,300 M^{-1} cm^{-1}\)

\(0.765=86,300 M^{-1}cm^{-1}\times C\times 1.3 cm\)

\(C=6.82\times 10^{-6} M=6.82\times10^{-6} mol/L\)

\(\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6\)

Both the masses are in grams.

Mass of solute that is chlorophyll = \(6.82\times 10^{-6}\times 893.49 g/mol=0.00609 g\)

Mass of ethanol = m

Density of the ethanol = d = 0.785 g/mL

Volume of the ethanol = 1 L = 1000 mL

\(m=0.785 g/mL\times 1000 mL=785 g\)

Mass of solution = 0.00609 g + 785 g = 785.00609 g

Concentration in ppm:

\(\frac{0.00609 g}{785.00609 g}\times 10^6=7.76\)

7.76 is the concentration of the solution in ppm.

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