Question

Ergosterol, a precursor of vitamin D, has δmax = 282 nm and molar absorptivity ε = 11,900 M-1cm-1. What is the concentration of ergosterol in a solution whose absorbance A = 0.93 with a sample pathlength l = 1.00 cm. M. What is the absorbance of a solution having a concentration of ergosterol of 4.1×10-5 M with a sample pathlength l = 1.00 cm?

Answer 1

The concentration of ergosterol in a solution can be calculated using Beer's law, which states that the absorbance of a sample is directly proportional to the concentration of the compound and the path length of the sample. Using the given values, the concentration of ergosterol in the solution is approximately 7.8 x 10^-5 M. The absorbance of a solution with a concentration of ergosterol of 4.1 x 10^-5 M is approximately 0.49.

Step-by-step explanation:

The concentration of ergosterol in a solution can be calculated using Beer's law. Beer's law states that the absorbance of a sample is directly proportional to the concentration of the compound and the path length of the sample.

To calculate the concentration of ergosterol in the solution with an absorbance of 0.93 and a path length of 1.00 cm, we can rearrange the equation to solve for concentration: A = εcl (where A is absorbance, ε is molar absorptivity, c is concentration, and l is the path length).

Using the given values, the concentration of ergosterol in the solution is calculated to be approximately 7.8 x 10^-5 M.

To calculate the absorbance of a solution with a concentration of ergosterol of 4.1 x 10^-5 M and a path length of 1.00 cm, we can use the same equation to solve for absorbance: A = εcl.

Using the given values, the absorbance of the solution is calculated to be approximately 0.49.

Answer 2

1)
`7.8* 10^(-5) M` is the concentration of ergosterol in a solution.

2) The absorbance of the solution is 0.49.

Step-by-step explanation:

Using Beer-Lambert's law :

Formula used :

`A=\epsilon * C* l`

where,

A = absorbance of solution

C = concentration of solution =
`2.00* 10^(-3)M`

l = path length = 1.00 cm

`\epsilon` = molar absorptivity coefficient

1)

We have:

Absorbance of the solution ,A = 0.93

Concentration of ergosterol ,C = ?

Path length ,l = 1.00 cm

`\epsilon = 11,900 M^(-1) cm^(-1)`

`A=\epsilon * C* l`

`C=(A)/(\epsilon* l)`

`=(0.93)/(11,900 M^(-1)cm^(-1)* 1.00 cm)=7.8* 10^(-5) M`

`7.8* 10^(-5) M` is the concentration of ergosterol in a solution.

2)

We have:

Absorbance of the solution ,A = ?

Concentration of ergosterol ,C =
`4.1* 10^(-5) M`

Path length ,l = 1.00 cm

`\epsilon = 11,900 M^(-1) cm^(-1)`

`A=\epsilon * C* l`

`=11,900 M^(-1)cm^(-1)* 4.1* 10^(-5)M* 1.00 cm= 0.49`

The absorbance of the solution is 0.49.