A solution was prepared by dissolving 118.9 mg of K2Cr2O7 in 1.000 L of water. A 10.00 mL aliquot was transferred to a 100.0 mL volumetric flask and diluted to the mark. The dilute solution had an absorbance of 0.206 at 400 nm, in a 1.00 cm cell. a. What is the concentration of the dilute solution in ppm? b. Using Beer’s Law equation, calculate the molar absorptivity of K2Cr2O7

a. To calculate the concentration of the dilute solution in ppm, we first need to find the concentration in molarity.

The molar mass of K2Cr2O7 is approximately 294.18 g/mol.

So, the initial concentration of the solution is (118.9 mg / 294.18 g/mol) / 1.000 L = 0.000404 mol/L or 0.404 mM.

Then, a 10.00 mL aliquot of this solution was diluted to 100.0 mL, so the final concentration is 0.404 mM * (10.00 mL / 100.0 mL) = 0.0404 mM.

To convert this to ppm, we use the fact that 1 ppm = 1 mg/L.

So, the concentration in ppm is 0.0404 mM * (294.18 mg / 1 mol) = 11.88 ppm.

b. Beer's Law is given by A = εlc, where A is the absorbance, ε is the molar absorptivity, l is the path length, and c is the concentration.

We can rearrange this to find ε = A / (lc).

The absorbance A is given as 0.206, the path length l is 1.00 cm, and the concentration c is 0.0404 mM or 0.0404 mmol/L.

So, the molar absorptivity ε is 0.206 / (1.00 cm * 0.0404 mmol/L) = 5.10 L/(mmol*cm).