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Nezavi [6.7K]
4 years ago
15

In a laboratory experiment, you are asked to determine the molar concentration of a solution of an unknown compound, X. The solu

tion 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?
Chemistry
1 answer:
bixtya [17]4 years ago
4 0

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

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