What is the path length in centimeters when the transmittance is $0.692$ $0.692$ , the molar absorptivity is $7.39 \times 10^{4} L/(mol.cm)$ $7.39 xx 10^4 "L/(mol.cm)"$ and the concentration is $0.0000923 M$ $0.0000923 M$ ?

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What is the path length in centimeters when the transmittance is $0.692$ $0.692$ , the molar absorptivity is $7.39 \times 10^{4} L/(mol.cm)$ $7.39 xx 10^4 "L/(mol.cm)"$ and the concentration is $0.0000923 M$ $0.0000923 M$ ?

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Answer 1 · 2017-03-27T21:36:12

$0.101 cm$ $"0.101 cm"$

Explanation:

Beer's Law:

$absorbance = (molar absorptivity) \times (path length) \times (concentration)$ $"Absorbance" = ("molar absorptivity") xx ("path length") xx ("concentration")$

or

$A = e \times L \times c$ $A = e xx Lxx c$

So

$L = \frac{A}{e c}$ $L = A/ (e c)$

$L = \frac{0.692}{(7.39 \cdot 10^{4} \frac{L}{m o l \cdot c m} \cdot (9.23 \cdot 10^{- 5} \frac{m o l}{L}))}$ $L = 0.692 / ((7.39*10^4L/(mol*cm) * (9.23*10^-5 (mol)/L))$

$L = 0.101 cm$ $L = "0.101 cm"$

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What is the path length in centimeters when the transmittance is $0.692$ $0.692$ , the molar absorptivity is $7.39 \times 10^{4} L/(mol.cm)$ $7.39 xx 10^4 "L/(mol.cm)"$ and the concentration is $0.0000923 M$ $0.0000923 M$ ?

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What is the path length in centimeters when the transmittance is 0.6920.692, the molar absorptivity is 7.39×104L/(mol.cm)7.39 xx 10^4 "L/(mol.cm)" and the concentration is 0.0000923M0.0000923 M?

1 Answer

Explanation:

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What is the path length in centimeters when the transmittance is $0.692$ $0.692$ , the molar absorptivity is $7.39 \times 10^{4} L/(mol.cm)$ $7.39 xx 10^4 "L/(mol.cm)"$ and the concentration is $0.0000923 M$ $0.0000923 M$ ?