Question

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?

Answer 1

7.76 is the concentration of the solution in ppm.

Step-by-step explanation:

Using Beer-Lambert's law :

Formula used :

`A=\epsilon * C* 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)* C* 1.3 cm`

`C=6.82* 10^(-6) M=6.82*10^(-6) mol/L`

`\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}* 10^6`

Both the masses are in grams.

Mass of solute that is chlorophyll =
`6.82* 10^(-6)* 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* 1000 mL=785 g`

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

Concentration in ppm:

`(0.00609 g)/(785.00609 g)* 10^6=7.76`

7.76 is the concentration of the solution in ppm.

Answer 2

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