Question

The absorbance of a crystal violet solution in water (density = 1.00 g/mL) is measured in a 1.0 cm cuvette. The absorbance is 0.614 and the molar absorptivity coefficient is 87,000 M–1cm–1. The molecular weight of crystal violet is 407.98 g/mol.

a. What is the molar concentration of the solution (M)?

b. How much crystal violet (in mg) was dissolved to prepare the 2.00 mL sample that was measured in the cuvette?

c.What is the concentration of the solution in ppm?

Answer 1

`7.0575* 10^(-6) M` is the molar concentration of the solution.

5.759 mg of crystal violet was dissolved to prepare the 2.00 mL sample that was measured in the cuvette.

Concentration of the solution in ppm is 2,879.31.

Step-by-step explanation:

Using Beer-Lambert's law :

Formula used :

`A=\epsilon * C* l`

where,

A = absorbance of solution

C = concentration of solution

`\epsilon` = The molar absorptivity coefficient

We have :

A = 0.614 , l = 1.0 cm , C= ?

`\epsilon =87,000 M^(-1) cm^(-1)`

`C=(A)/(\epsilon l)=(0.614 )/(87,000 M^(-1) cm^(-1)* 1 cm)`

`C=7.0575* 10^(-6) M`

`7.0575* 10^(-6) M` is the molar concentration of the solution.

`Moles (n)=Molarity(M)* Volume (L)`

Moles of crystal violet = n

Volume of crystal violet solution = 2.00 mL = 0.002 L

Molarity of the crystal violet =
`C=7.0575* 10^(-6) M`

`n=7.0575* 10^(-6) M* 0.002 L=1.4115* 10^(-5) mol`

Mass of
`1.4115* 10^(-5) mol` of crystal violet:

`1.4115* 10^(-5) mol* 407.98 g/mol=0.005759 g`

0.005759 g = 5.759 mg (1 g = 1000 mg)

5.759 mg of crystal violet was dissolved to prepare the 2.00 mL sample that was measured in the cuvette.

`ppm = \frac{\text{mass of compound (mg)}}{\text{volume of solution}L}`

Mass of crystal violet = 5.759 mg

Volume of solution= 2 mL = 0.002 L

Concentration of the solution in ppm:

`=(5.759 mg)/(0.002 L)=2879.31 ppm`