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?

Let A= ebc
0.765 = 86,300*1.3*c

Solve for c = approximately 7E-6 Molar
= mols/L soln.
g = 7E-6*893.49
= 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 Link

Tringa0 Tringa0 · Answer 2 · 2020-04-08T09:50:20+02:00

Answer:

7.76 is the concentration of the solution in ppm.

Explanation:

Using Beer-Lambert's law :

Formula used :

[tex]A=\epsilon \times C\times l[/tex]

where,

A = absorbance of solution = 0.765

C = concentration of solution = ?

l = path length = 1.3 cm

[tex]\epsilon[/tex] = molar absorptivity coefficient = [tex]86,300 M^{-1} cm^{-1}[/tex]

[tex]0.765=86,300 M^{-1}cm^{-1}\times C\times 1.3 cm[/tex]

[tex]C=6.82\times 10^{-6} M=6.82\times10^{-6} mol/L[/tex]

[tex]\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6[/tex]

Both the masses are in grams.

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

Mass of ethanol = m

Density of the ethanol = d = 0.785 g/mL

Volume of the ethanol = 1 L = 1000 mL

[tex]m=0.785 g/mL\times 1000 mL=785 g[/tex]

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

Concentration in ppm:

[tex]\frac{0.00609 g}{785.00609 g}\times 10^6=7.76[/tex]

7.76 is the concentration of the solution in ppm.