Calculating Concentration using Beer-Lambert Law
This problem requires calculating the concentration of a compound using the Beer-Lambert Law, given absorbance, path length, and molar absorptivity.
Beer-Lambert Law Principle
The Beer-Lambert Law relates the attenuation of light to the properties of the material through which the light is traveling. It is stated as:
$ A = \varepsilon \times C \times l $
Where:
- \( A \) is the absorbance (unitless).
- \( \varepsilon \) is the molar absorptivity (in \( M^{-1} \ cm^{-1} \)).
- \( C \) is the concentration (in \( M \)).
- \( l \) is the path length (in \( cm \)).
Given Information
- Absorbance, \( A = 0.42 \)
- Wavelength, \( \lambda = 275 \ nm \) (Note: Wavelength is not directly used in the calculation but specifies the condition for \( \varepsilon \))
- Path length, \( l = 0.1 \ dm \)
- Molar absorptivity, \( \varepsilon_{275} = 8.4 \times 10^3 \ M^{-1} \ cm^{-1} \)
Unit Conversion for Path Length
The molar absorptivity is given in \( cm^{-1} \), so the path length must be converted from decimeters (dm) to centimeters (cm).
- \( 1 \ dm = 10 \ cm \)
- Therefore, \( l = 0.1 \ dm \times 10 \ cm/dm = 1 \ cm \)
Calculation of Concentration
Rearranging the Beer-Lambert Law equation to solve for concentration \( C \):
$ C = \frac{A}{\varepsilon \times l} $
Substitute the known values:
$ C = \frac{0.42}{(8.4 \times 10^3 \ M^{-1} \ cm^{-1}) \times (1 \ cm)} $
$ C = \frac{0.42}{8.4 \times 10^3} \ M $
$ C = \frac{42 \times 10^{-2}}{8.4 \times 10^3} \ M $
$ C = \left( \frac{42}{8.4} \right) \times 10^{-2-3} \ M $
$ C = 5 \times 10^{-5} \ M $
Final Answer
The concentration of the compound is \( 5 \times 10^{-5} \ M \). The question asks for the value to fill in the blank as \( \_\_\_\_\_\_\_\_\_\_ \times 10^{-5} \ M \). Therefore, the required value is 5.
The value is 5.
Important Questions from Enzyme Assays Molar Extinction Coefficient
- A solution shows a transmittance of 20% when taken in a cuvette of 2.5 cm path length. If the molar absorption coefficient of the solution is $12000 \text{ dm}^3/\text{mol.cm}$, the concentration of the solution is ________ $\times 10^5 \text{ mol/dm}^3$ (rounded off to two decimal places).
- A solution containing GTP has molar extinction coefficient of $1.55 \times 10^4$ $mol^{-1}dm^3cm^{-1}$ at a given wavelength. The concentration of GTP solution is $1.290 \times 10^{-5}$ $mol$ $dm^{-3}$. The absorbance of GTP solution in 1 cm cuvette at the same wavelength will be .................
- An enzyme preparation has activity of 2 Units per 20 $\mu$l, and protein concentration 0.4 mg/ml. The specific activity (Units/mg) of this enzyme will be ________
- Measurement of the absorbance of a solution containing NADH in a path length of 1cm cuvette at 340 nm shows the value of 0.31. The molar extinction coefficient of NADH is $6200 M^{-1} cm^{-1}$. The concentration of NADH in the solution is ________ $\mu M$ (correct to integer number).
- If a $10$ mM solution of a biomolecule in a cuvette of path length $10$ mm absorbs $90\%$ of the incident light at $280$ nm, the molar extinction coefficient of the biomolecule at this wavelength is ________ $M^{-1}cm^{-1}$. (Round off to two decimal places)
