Question

if you had a dye with a molecular weight of 574.76 amu which has a molar absorptivity of 82,000 at 550 nm, what would be the expected value of absorbance at that wavelength if the dye was present at a concentration of 1.0 mg/L?

Answer 1

Answer on the question #59723, Chemistry / General Chemistry

Question:

if you had a dye with a molecular weight of 574.76 amu which has a molar absorptivity of 82,000 at $550\mathrm{nm}$, what would be the expected value of absorbance at that wavelength if the dye was present at a concentration of $1.0\mathrm{mg/L}$?

Solution:

According to Beer-Lambert law the absorbance $A$ is the product of molar absorptivity $\varepsilon$, concentration of the dye $c$ and length of the spectroscopic cell $l$:

$$A = \varepsilon \cdot l \cdot c$$

If we assume 1cm length of spectroscopic cell:

$$A = \varepsilon \cdot c$$

Molar concentration of the dye is (the value of molar mass in $g \cdot mol^{-1}$ is equal to the value of molecular mass in amu):

$$c = \frac{m}{M} = \frac{1.0(mg \cdot L^{-1})}{574.76(g \cdot mol^{-1})} = 1.74 \cdot 10^{-6}mol \cdot L^{-1}$$ $$A = 82000(L \cdot mol^{-1}) \cdot 1.74 \cdot 10^{-6}(mol \cdot L^{-1}) = 0.143$$

Answer: Expected absorbance is 0.143.

http://www.AssignmentExpert.com/