Concentration problem Please help

Question

An unknown amount of a compound with a molecular mass of 259.36 g/mol is dissolved in a 10 mL volumetric flask. A 1.00 mL aliquot of this solution is transferred to a 25 mL volumetric flask, and enough water is added to dilute to the mark. The absorbance of this diluted solution at 337 nm is 0.4120 in a 1.000 cm cuvette. The molar absorptivity for this compound at 337 nm337 nm is 𝜖337=5573 M−1cm.

What is the concentration of the compound in the cuvette?

What is the concentration of the compound in the 10 mL flask?

How many milligrams of the compound was used to make the 10 mL solution?

Answer 1

Beer-Lambert Law:

A = 𝜖Cl

A = absorbance = 0.4120

𝜖 = extinction coefficient = molar absorptivity = 5573 M/cm

C = concentration (M)

l = path length = 1 cm

Solving for C, we have...

0.4120 = (5573) (C) (1)

C = 7.39x10^-5 M = Concentration in the cuvette

The concentration in the cuvette represents is the same as the concentration in the 25 ml flask, and this is a 1 to 25 (25 fold) dilution of the original solution. Therefore..

7.39x10^-5 M x 25 = 1.84x10^-3 M = concentration in the 10 ml flask

To find the mg of substance in the 10 ml flask, we can use the molar mass of the substance (259.35 g/mole) and the concentration, as follows:

1.84x10^-3 mol / L x 259.35 g / mole = 0.4772 g / L x 0.01 L x 1000 mg / g = 4.77 mg in 10 mls (0.01L)

(be sure to check the math)