An unknown amount of a compound with a molecular mass of 273.67 g/mol is dissolv

Question

An unknown amount of a compound with a molecular mass of 273.67 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 351 nm is 0.545 in a 1.000-cm cuvette. The molar absorptivity for this compound at 351 nm is 351 = 6083 M–1 cm–1. (a) What is the concentration of the compound in the cuvete? (b) What is the concentration of the compound in the 10-mL flask? (c) How many mg of compound were used to make the 10-mL solution?

Explanation / Answer

Apply Beer's Law

A = e*l*C

A = absorbanc, e = molar absorptivity, l = length of cuvette

so

from the cuvette concentation

A = e*l*C

C = A/(e*l) = 0.545/(1*6083 ) = 0.00008959M

this value is A) concentration of compound in the cuvetet

b)

concentration of compound in th 10 mL flask:

0.00008959 M at 25 mL

find concentation at V = 1 mL

M1*V1 = M2*V2

M1*1 = 0.00008959*25

M1 = 0.00223975 M @1 mL

note that the concentration at 1 mL is the same of that in V = 10 mL

so this is B) M1 = 0.00223975 M

c)

find mass in mg for compound

moles = C*V = 0.00223975 M * 10 mL= 0.0223975 mmol

1 mmol = 273.67 mg

0.0223975 mg = 0.0223975 *273.67 = 6.129 mg of sample

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