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`

Step-by-step 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 = (A)/( \epsilon * L )`

=>
`C_2 = (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`