Calculating Proline Concentration Using Beer-Lambert Law
This question requires calculating the concentration of a proline ninhydrin complex using the Beer-Lambert Law.
Beer-Lambert Law Principle
The Beer-Lambert Law relates the absorbance of a solution to the concentration of the analyte and the path length the light travels through the solution. The formula is:
$A = \epsilon lc$
Where:
- A is the absorbance (unitless)
- \(\epsilon\) is the molar extinction coefficient (in $\mu$M$^{-1}$Cm$^{-1}$)
- l is the path length of the cuvette (in cm)
- C is the concentration of the analyte (in $\mu$M)
Given Information
- Absorbance, A = 0.68
- Molar extinction coefficient, \(\epsilon\) = 0.34 $\mu$M$^{-1}$Cm$^{-1}$
- Path length, l = 1 cm
Calculation Steps
- Rearrange the Beer-Lambert Law formula to solve for concentration (C):
$C = \frac{A}{\epsilon l}$
- Substitute the given values into the rearranged formula:
$C = \frac{0.68}{0.34 \, \mu \text{M}^{-1}\text{Cm}^{-1} \times 1 \, \text{cm}}$
- Calculate the concentration:
$C = \frac{0.68}{0.34} \, \mu \text{M}$
$C = 2 \, \mu \text{M}$
Result
The concentration of the proline is 2 $\mu$M. This corresponds to Option 4.
Important Questions from Chemical Equilibrium
- The amount of a particular gas dissolved in water depends on
(i) its solubility in water.
(ii) its partial pressure at the air/water interface or sediment/water interface.
(iii) the water temperature.
(iv) the levels of salts in the water.
Identify the correct answer : - A 50 ml solution of pH = 1 is mixed with a 50 ml solution of pH = 2. The pH of the mixture will be nearly
- The solubility product of the following type of reaction :
$Al(OH)_3 \rightarrow Al^{3+} + 3\bar{O}H$ is Consider the reaction $2C(s) + O_2(g) \rightleftharpoons 2CO(g)$ at equilibrium.
The equilibrium can be shifted towards the forward direction by- Consider the reaction:
$A+B \rightleftharpoons C$
The unit of the thermodynamic equilibrium constant for the reaction is
