Question

If the molar extinction coefficient (e) of a compound is 7,500L mol'cm' at 443 nm and the Abs (at 433 nm) = 0,89 find the concentration of that solution. (assume a path length of 1.0 cm)

Answer 1

To find the concentration of the solution using the molar extinction coefficient and absorbance, apply Beer's Law. With the given absorbance of 0.89 and molar extinction coefficient of 7,500 L mol-1cm-1, the concentration is calculated to be 1.19 × 10-4 mol L-1.

Step-by-step explanation:

You've asked how to find the concentration of a solution using the molar extinction coefficient, absorbance, and path length. This is possible through the application of Beer's Law, which is expressed as A = εcL, where A is the absorbance, ε is the molar extinction coefficient, c is the concentration, and L is the path length. Given that the molar extinction coefficient (e) is 7,500 L mol-1cm-1 at 443 nm, the absorbance (Abs) at 433 nm is 0.89, and assuming a path length of 1.0 cm, we can rearrange Beer's Law to solve for concentration (c): c = A / (εL).

Substituting the given values:

c = 0.89 / (7,500 L mol-1cm-1 × 1 cm)

c = 0.89 / 7,500 mol L-1

Therefore, the concentration of the solution is 1.19 × 10-4 mol L-1 or 119 × 10-6 M.