A compound with a molecular weight of 292.16 was dissolved in a 5-ml volumetric

Question

A compound with a molecular weight of 292.16 was dissolved in a 5-ml volumetric flask. A 1.00-mLaliquat was withdrawn, placed in a 10-mL volumetric flask, and diluted to the mark. The absorbance measured at 340 nm was 0.427 in a 1.000-cm cuvette. The molar absorptivity for this compound at 340 nmis 340-6130 M-1cm', a) Calculate the concentration of the compound in the cuvette b) What was the concentration of compound in the 5-mL flask c) How many milligrams of compound were used to make the 5-mL solution?

Explanation / Answer

The relation of absorbance of a substance at a particular wavelength with the path length of light through the sample and the sample concentration is given by the Beer-Lambert law as Absorbance = Molar absorptivity*concentration*path length with concentration in molarity and path length in cm.

a) Thus, for the given substance, concentration in molarity is given as c = Absorbance/(alpha*path length) = 0.427/(6130*1) = 6.9657x10-5mol/L.

b) Since 1mL of the inital 5mL solution was diluted to 10mL, the concentration in 1mL of the stock solution is given from the law of equivalence as 6.9657x10-4mol/L. This gives the concentration in the stock 5mL solution to be 5*6.9657x10-4mol/L =.34.8285x10-4mol/L.

c) Since the given concentration gives the moles per liter of solution, the actual moles in 5mL is gives as (34.8285x10-4mol/L/1000)*5 = 1.7414x10-5mol. Now using the molar mass of the compound as 292.16g/mol, the mass of compound in the 5mL solution is given as 1.7414x10-5mol*292.16g/mol = 5.0876mg.

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