Question

One milligram of a compound of molecular weight 160 is dissolved in 10 mL of ethanol, and the solution is poured into a 1-cm UV cell. The UV spectrum is taken, and there is an absorption at ????max = 247 nm. The maximum absorbance at 247 nm is 0.50. Calculate the value of ???? for this absorption.

Answer 1

To calculate the value of ɛ using the Beer-Lambert Law, determine the molarity of the compound in the solution and divide the absorbance by the product of concentration and path length. For the given problem, the calculated ɛ value is 80,000 L/mol*cm.

Step-by-step explanation:

To calculate the value of ɛ (molar absorptivity or extinction coefficient) for the given absorption, we can use the Beer-Lambert Law, which relates absorbance (A) to ɛ, the concentration of the solution (c), and the path length (l), expressed as A = ɛcl. In the provided problem, we are told that the maximum absorbance (Amax) at 247 nm is 0.50 in a 1-cm pathlength UV cell. Given that we have 1 mg of the compound dissolved in 10 mL of ethanol, we first need to calculate the molarity of the solution, which is the concentration in moles/liter (mol/L). The molar mass of the compound is 160 g/mol, so the concentration c = (0.001g/160g/mol)/(0.01L) = 6.25 x 10-6 mol/L. With a known path length of 1 cm (path length = 1 cm), we can rearrange the Beer-Lambert Law to solve for ɛ: ɛ = A/(cl). Plugging in the values, we obtain ɛ = 0.50/(6.25 x 10-6 mol/L * 1 cm) = 80,000 L/mol*cm.

Answer 2

The value of molar absorptivity of the compound is
`800 L mol^(-1) cm^(-1)`.

Step-by-step explanation:

Using Beer-Lambert's law :

`A=\epsilon * C* l`

A = absorbance of solution

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

l = path length = 1.00 cm

`\epsilon` = molar absorptivity coefficient

`Concentration (C)= \frac{\text{moles of compound}}{\text{Volume of the solution(L)}}`

Moles of compound =
`(1 mg)/(160 g/mol)=(0.001 g)/(160 g/mol)=6.25* 10^(-6) mol`

Volume of the solution = 10 mL = 0.010 L

`C=(6.25* 10^(-6) mol)/(0.010 L)=0.000625 mol/L`

A = 0.50 , l = 1 cm

`0.50=\epsilon * 0.000625 mol/L* 1 cm`

`\epsilon =(0.50)/(0.000625 mol/L* 1 cm)=800 L mol^(-1) cm^(-1)`

The value of molar absorptivity of the compound is
`800 L mol^(-1) cm^(-1)`.