Solve 0.1% w/w Concentration Problem: Beer's Law & C15H6N4

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SUMMARY

The discussion focuses on calculating the amount of a C15H6N4 indicator needed to achieve a 0.1% w/w concentration in water, given a transmittance of 0.35 and an extinction coefficient (ε) of 8.79 x 10^5 at a wavelength (λmax) of 580 nm. Using Beer's Law (A = εbc), the absorbance (A) was calculated as 0.456, leading to a concentration of 5.08 x 10^-7 mol/L. The amount of indicator required depends on the final volume of the solution, which was not specified in the discussion.

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monae
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Problem:
Assume the human eye can detect light with transmittance of 0.35.
The colored form of the indicator has ε = 8.79 x 10^5. and λmax = 580 nm. How
much indicator in ml should be used if the indicator has concentration 0.1% w/w in water.

Equation Used:
Beer's Law = A = εbc
where ε is given as 8.79 x 10^5
b = 1.00cm
A= -logT = -log(0.35) = 0.456

Using Beer's Law, I solved for C and got 5.08(10^-7)mol/L
But I don't know what to do to find the amount of indicator needed.
The molecular formula is C15H6N4 if that helps.
 
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Looks like you are going to need 5.08 X 10^-7 moles per liter to see it. How much of the 0.1% w/w of the indicator (you should probably convert this to mol/L using the molecular formula) will you need?

That depends on how much solution you will have to dilute this into (what is the volume of the cell). This information is missing in your problem.
 

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