Question

In a laboratory experiment, you are asked to determine the molar concentration of a solution of an unknown compound, X. The solution diluted in with water (200 µL of X + 800 µL of H2O) has an absorbance at 425 nm of 0.8 and a molar extinction coefficient of 1.5 x103 M-1cm-1 at 425 nm. What is the molar concentration of the original solution of X?

Answer 1

Complete Question

In a laboratory experiment, you are asked to determine the molar concentration of a solution of an unknown compound, X. The solution diluted in with water (200 µL of X + 800 µL of H2O) has an absorbance at 425 nm of 0.8 and a molar extinction coefficient of 1.5 x103 M-1cm-1 at 425 nm. What is the molar concentration of the original solution of X? (1 cm cuvette)

Answer:

The original concentration is $C_1 = 0.0027 M$

Explanation:

From the question we are told that

The original volume of solution X is $V_1 = 200 \mu L$

The volume of solution X after dilution is $V_ = 200 + 800 = 1000 \mu L$

The absorbance is $A = 0.8$

The molar extinction coefficient is $\epsilon = 1.5 *10^{3} \ M^{-1} cm^{-1}$

Generally from Beer's law

$A = \epsilon * C * L$

Here

L is the path length with a value of 1 cm

C_2 is the concentration of the solution at the given absorbance

=> $C_2 = \frac{A}{ \epsilon * L }$

=> $C_2 = \frac{0.8}{1.5 *10^{3} * 1 }$

=> $C_ = 5.33*10^{-4} \ M$

Generally we have that

$C_1 *200 = 5.33*10^{-4} * 1000$

=> $C_1 = 0.0027 M$