A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (ε) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture?| Wavelength (λ, nm) | Molar Absorptivity (ε, M⁻¹cm⁻¹) || ------------------ | ------------------------------ || | X | Y || 443 | 15790 | 3725 || 520 | 3841 | 6.180×10³ |In this setup, Beer's Law can be used to calculate the concentrations of compounds X and Y in the mixture. Beer's Law states: \[ A = ε \cdot c \cdot l \] where:- \( A \) is the absorbance,- \( ε \) is the molar absorptivity,- \( c \) is the concentration,- \( l \) is the path length of the cell (1.00 cm in this case).The information provided will help determine the concentrations of X and Y based on their unique absorbance at given wavelengths.

According to Beer Lambert's Law, A=ε.C.l where A is the absorption, ε is the molar absorptivity, C is the concentration and l is the path length. Absorption is additive in nature.Write the Beer lambert's expression for the mixture at 443 nm. A443=εxCxl +εyCyl Substitute 0.584 for A443, 15790 M-1.cm-1 for εx and 3725 M-1.cm-1 for εy and 1 cm for l and rewrite the expression. 0.584=(15790 M-1.cm-1)Cx(1 cm) +(3725 M-1.cm-1)Cy(1 cm) Simplify the equation. 0.584=(15790 M-1.Cx) +(3725 M-1.Cy) Substitute 0.501 for A520, 3841 M-1.cm-1 for εx and 6.18×103 M-1.cm-1 for εy and 1 cm for l and rewrite the expression. 0.501=(3841 M-1.Cx) +(6.18×103 M-1.Cy) Solve for Cx and Cy. Cx=2.09×10-5 MCy=6.81×10-5 M

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A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (E) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture? Wavelengt h (A, nm) Molar Absorptivity (E, M-'cm-1) Y 443 15790 3725 520 3841 6.180x103

A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (E) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture? Wavelengt h (A, nm) Molar Absorptivity (E, M-'cm-1) Y 443 15790 3725 520 3841 6.180x103

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Concept explainers

UV Visible Spectroscopy

Spectroscopy refers to the absorption or emission of light or other radiations by matter. When a substance or chemical absorbs light or radiation, they reach an excited state followed by de-excitation or emission of the light or radiation forming distinct spectra. UV-visible spectroscopy or ultraviolet-visible refers to the absorption and emission in the ultraviolet and visible regions of the electromagnetic spectrum.

Question

A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (ε) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture?

| Wavelength (λ, nm) | Molar Absorptivity (ε, M⁻¹cm⁻¹) |
| ------------------ | ------------------------------ |
| | X | Y |
| 443 | 15790 | 3725 |
| 520 | 3841 | 6.180×10³ |

In this setup, Beer's Law can be used to calculate the concentrations of compounds X and Y in the mixture. Beer's Law states:
\[ A = ε \cdot c \cdot l \]
where:
- \( A \) is the absorbance,
- \( ε \) is the molar absorptivity,
- \( c \) is the concentration,
- \( l \) is the path length of the cell (1.00 cm in this case).

The information provided will help determine the concentrations of X and Y based on their unique absorbance at given wavelengths.

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