A sample of 0.300 mg pure chromium was added to excess hydrochloric acid to form a 10.0 mL aqueous solution of a chromium (III)

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A sample of 0.300 mg pure chromium was added to excess hydrochloric acid to form a 10.0 mL aqueous solution of a chromium (III)

salt, which has a violet hue. Exactly 1.00 mL of the resulting solution was analyzed using a spectrophotometer in a 1.00-cm cell at 575 nm, and the percent transmittance for the solution was 62.5%. What is the extinction coefficient?

Chemistry

1 answer:

eduard [2.7K] · Answer 1 · 2026-04-07T09:05:31+03:00

extinction coefficient (ε) = 347 L·mol⁻¹·cm⁻¹. The chemical equation representing the reaction of chromium (Cr) with hydrochloric acid (HCl) is: 2 Cr + 6 HCl → 2 CrCl₃ + 3 H₂. To find the number of moles, we apply the formula: number of moles = mass / molar weight. For chromium, we calculate: number of moles of Cr = 0.3 × 10⁻³ (g) / 52 (g/mole), leading to number of moles of Cr = 5.77 × 10⁻⁶ moles. Examining the reaction, we observe that 2 moles of Cr yield 2 moles of CrCl₃, hence 5.77 × 10⁻⁶ moles of Cr will also produce 5.77 × 10⁻⁶ moles of CrCl₃. The molar concentration is determined by: molar concentration = number of moles / volume (L), thus molar concentration of CrCl₃ = 5.77 × 10⁻⁶ / 10 × 10⁻³, which equals 5.77 × 10⁻⁴ moles/L. To convert percent transmittance (%T) to absorbance (A), we use the equation A = 2 - log(%T). Therefore, A = 2 - log(62.5), leading to A = 0.2. The relationship defining absorbance (A) includes the extinction coefficient (ε), path length (l), and concentration (c): A = εlc, hence ε = A / lc, giving ε = 0.2 / (1 × 5.77 × 10⁻⁴), which results in ε = 0.0347 × 10⁴. Thus, the extinction coefficient is ε = 347 L·mol⁻¹·cm⁻¹.