Question

A solution containing 3.8mg/100ml of compound B whose molecular weight is 280g has a trasmittance of 39.6% in a 1cm cell at 480nm. Calculate molar absorptivity of compound B?

Answer 1

The molar absorptivity is `"300 m"^2"mol"^"-1"`.

Explanation:

The transmittance `T` of a sample is defined as

`color(blue)(bar(ul(|color(white)(a/a)T = I/I_0color(white)(a/a)|)))" "`

where `I_0` and `I` represent the intensity of the radiation entering and leaving the sample.

It is convenient to define the absorbance `A` as

`color(blue)(bar(ul(|color(white)(a/a)A = "-log"Tcolor(white)(a/a)|)))" "`

The formula for Beer's Law is

`color(blue)(bar(ul(|color(white)(a/a)A = epsilonclcolor(white)(a/a)|)))" "`

where

`epsilon color(white)(l) =` the molar absorptivity
`c color(white)(l) =` the molar concentration
`l color(white)(l)=` the path length of the cell

We can rearrange this formula to get

`epsilon = A/(cl)`

So,

`epsilon = ("-log"T)/(cl)`

In your problem,

`T = 0.396`

`c color(white)(l)= (0.0038 color(red)(cancel(color(black)("g"))))/("0.100 dm"^3) × ("1 mol")/(280 color(red)(cancel(color(black)("g")))) = 1.36 × 10^"-4"color(white)(l)"mol·dm"^"-3" = "0.136 mol·m"^"-3"`

`l color(white)(l)= "1 cm" = "0.01 m"`

∴ `epsilon = ("-log"(0.396))/("0.136 mol·m"^"-3" × "0.01 m") = "300 m"^2"mol"^"-1"`