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Chapter 17

Q. 17.3

Measuring Benzene in Hexane

(a)  Pure hexane has negligible ultraviolet absorbance above a wavelength of 200 nm. A solution prepared by dissolving 25.8 mg of benzene (C_{6}H_{6}, FM 78.11) in hexane and diluting to 250.0 mL had an absorption peak at 256 nm and an absorbance of 0.266 in a 1.000-cm cell. Find the molar absorptivity of benzene at this wavelength.

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Verified Solution

The concentration of benzene is

[C_{6}H_{6}] = \frac{(0.025  8  g)/(78.11  g/mol)}{0.250  0  L} = 1.32_{1} × 10^{−3} M

We find the molar absorptivity from Beer’s law:

Molar   absorptivity = ε = \frac{A}{bc} = \frac{(0.266)}{(1.00  cm)(1.32_{1} × 10^{−3}  M)} = 201._{3}  M^{−1}  cm^{−1}

(b)  A sample of hexane contaminated with benzene had an absorbance of 0.070 at 256 nm in a cuvet with a 5.000-cm pathlength. Find the concentration of benzene in mg/L.

Using Beer’s law with the molar absorptivity from part (a), we find

[C_{6}H_{6}] = \frac{A}{εb} = \frac{0.070}{201._{3}  M^{−1}  cm^{−1} (5.00  cm)} = 6.9_{5} × 10^{−5} M

[C_{6}H_{6}] = (6.9_{5} × 10^{−5}  \frac{mol}{L})(78.11 × 10^{3}  \frac{mg}{mol}) = 5.4  \frac{mg}{L}

Test Yourself     0.10 mM KMnO_{4} has an absorbance maximum of 0.26 near 525 nm in a 1.000-cm cell. Find the molar absorptivity and the concentration of a solution whose absorbance is 0.52 at 525 nm in the same cell. (Answer: 2 600 M^{−1}  cm^{−1} , 0.20 mM)

This example illustrates the measurement of molar absorptivity from a single solution. It is better to measure several concentrations to obtain a more reliable absorptivity and to demonstrate that Beer’s law is obeyed.