Question

An unknown amount of a compound with a molecular mass of 264.37 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 327 nm is 0.487 in a 1.000 cm cuvette. The molar absorptivity for this compound at 327 nm is ϵ 327 = 6503 M^(−1) cm^(−1).
(a) What is the concentration of the compound in the cuvette?
(b) What is the concentration of the compound in the 10-mL flask?
(c) How many milligrams of the compound were used to make the 10-mL solution?

Answer 1

Answer:

a) 0.0019 M is the concentration of the compound in the 10-mL flask.

b) $7.488\times 10^{-5} M$ is the concentration of the compound in the cuvette.

c) 5.0 milligrams of the compound were used to make the 10-mL solution.

Explanation:

a)

Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution = 0.487

C = concentration of solution = ?

l = length of the cell = 1.000 cm

$\epsilon$ = molar absorptivity of this solution = $6503 M^{-1}cm^{-1}$

$0.487=6503 M^{-1}cm^{-1}\times C\times 1.000 cm$

$C=\frac{0.487}{6503 M^{-1} cm^{-1}}=7.488\times 10^{-5} M$

$7.488\times 10^{-5} M$ is the concentration of the compound in the cuvette.

b) Concentration of solution in 25 mL of volumetric flask = $C_1=7.488\times 10^{-5} M$

Volume of diluted solution = $V_1=25 mL$

Concentration of solution in 10 mL of volumetric flask = $C_2=?$

Volume of solution taken before dilution = $V_2=1 mL$

$C_1V_1=C_2V_2$

$C_2=\frac{C_1V_1}{V_2}=\frac{7.488\times 10^{-5} M\times 25 mL}{1 mL}$

$C_2=0.0019 M$

0.0019 M is the concentration of the compound in the 10-mL flask.

c)

Concentration of solution in 10 mL of volumetric flask = 0.0019 M

Moles of unknown compound = n

Volume of the solution = 10 mL = 0.010 L

1 mL = 0.001 L

$Concentration =\frac{Moles}{Volume (L)}$

$n=0.0019 M\times 0.010 L=0.000019 mol$

Molar mass of unknown compound = 264.37 g/mol

Mass of 0.00019 moles of unknown compound:

0.000019 mol × 264.37 g/mol = 0.0050 g

1 g = 1000 mg

0.050 g= 0.0050 × 1000 mg = 5.0 mg

5.0 milligrams of the compound were used to make the 10-mL solution.