A sample weighing 1.25 mg containing a certain compound M (FW 292.16 g/mol) was dissolved in

solvent in a 5.00 mL volumetric flask. A 1.00 mL aliquot was drawn up, placed in a 10.00 mL volumetric and
diluted to the mark with solvent. The absorbance of this solution at 340 nm was 0.427 in a 1.00 cm
cuvette. The molar absorptivity at 340 nm for the compound is 6130 M�]1 cm�]1.
(a) Calculate the concentration of compound M in the 5 mL volumetric flask.
(b) Calculate the percent purity of the compound.

Conc= mass/volume

=1.25/5
=0.25mg/ml

To calculate the concentration of compound M in the 5 mL volumetric flask, we can use the Beer-Lambert Law. The equation is given by:

A = εbc

Where:
A = Absorbance (0.427)
ε = Molar absorptivity (6130 M^-1 cm^-1)
b = Path length (1.00 cm)
c = Concentration

Rearranging the equation, we can solve for concentration (c):

c = A / (εb)

Substituting the given values:

c = 0.427 / (6130 M^-1 cm^-1 * 1.00 cm)

c ≈ 6.96 x 10^-5 M

Therefore, the concentration of compound M in the 5 mL volumetric flask is approximately 6.96 x 10^-5 M.

To calculate the percent purity of the compound:
The compound was dissolved in a 5 mL volumetric flask, and a 1 mL aliquot was taken for further dilution. The dilution factor for this aliquot is:

Dilution factor = Final volume / Initial volume
= 10 mL / 1 mL
= 10

Since the compound is diluted by a factor of 10, the concentration of the compound in the original sample can be calculated by dividing the concentration in the 5 mL volumetric flask by the dilution factor:

Concentration in the original sample = Concentration in 5 mL flask / Dilution factor
= 6.96 x 10^-5 M / 10
= 6.96 x 10^-6 M

Now, to calculate the percent purity, we need to compare the mass of the compound in the sample to the mass of the pure compound.

Mass of the compound in the sample = Concentration x Volume x Molar Mass
= 6.96 x 10^-6 M x 1.25 mg x (1 g / 1000 mg) x (1 mol / 292.16 g)
≈ 2.85 x 10^-10 mol

Mass of the pure compound = Concentration x Volume x Molar Mass
= 6.96 x 10^-6 M x 1 mL x (1 g / 1000 mg) x (1 mol / 292.16 g)
≈ 2.03 x 10^-9 mol

Percent purity = (Mass of the compound in the sample / Mass of the pure compound) x 100
= (2.85 x 10^-10 mol / 2.03 x 10^-9 mol) x 100
≈ 14.0%

Therefore, the percent purity of the compound is approximately 14.0%.