I have no idea What to do. What does it mean by 'draw what you think the visible range spectrum of this dye would look like.' To get the A, I used Beer's law. So I got 9.265nm for Absorbance. What should I do for the next?
The question is
----Next Simon checks the spectrophotometer. He takes out his traveling standard solution - a 2.50 x 10-5 M solution of a particularly stable dye It has a molar absorptivity of 3.85 x 105 at 585 nm. What value for absorbance at 585 nm does Simon expect to get? Draw what you think the visible range spectrum of this dye would look like.
How did you get 9.265 nm for absorbance? A nm is a unit of length, not absorbance?
For the next part,
A = k*c (I've omitted the cell length since that usually is a constant value).
A = molar absorptivity x concn(in M)
A = 3.85 x 10^5 x 2.50 x 10^-5 = ?? That is the absorbance Simon expects.That may be how you arrived at the 9.625 in the first part of your question. The drawing part is this.
Since the solution is being measured at 585 nm, one expects that will be the point at which A is a maximum. So you have a sheet of graph paper, label the Y axis as A units and the x axis as wavelength and show 400 nm in the left side and 700 nm on the right side. Then place a point at 585. That is where the maximum will be. Now draw a smooth curve, beginning at 400 nm and zero A to the point and back down to zero A at 700 nm (UNLESS you have some other information that gives different A values for other wavelengths. Here is a sample spectrum of how a visible spectrum should appear but the maximum is not at 585.
http://www.google.com/imgres?imgurl=http://www.uwplatt.edu/chemep/chem/chemscape/labdocs/catofp/measurea/concentr/spec20/pic/specq2.gif&imgrefurl=http://www.uwplatt.edu/chemep/chem/chemscape/labdocs/catofp/measurea/concentr/spec20/spec20w2.htm&h=288&w=384&sz=4&tbnid=-xoYNxe9NgbDlM:&tbnh=92&tbnw=123&prev=/images%3Fq%3Dabsorbance%2Bversus%2Bwavelength&zoom=1&q=absorbance+versus+wavelength&usg=__zFBGFRJTpFVBLcNYV9XkR26dij8=&sa=X&ei=zdOzTLnQJoOKlweis9yRDA&ved=0CB0Q9QEwAg
Then, What should be the unit for 9.625? I thought that is the absorbance. So I used nm unit which is for absorbance.
You need to work out the unit by including units in the calculation.
9.265 cannot be an absorbance as the scale on even the best instruments only goes to just above 3.
585 nm is a standard filter value so Simon dye sample may be such it has a maximum at 585 nm.
Absorbance itself does not have any units as it is a ratio of two log values. This is done by the instrument and is invisible to the operator. Only folks like me who built their own spectrometer had to do this by hand!
To determine the absorbance at 585 nm for the given dye, you first need to understand what the question is asking for. In this case, Simon is expecting to measure the absorbance of a dye solution at a specific wavelength, which is 585 nm.
The molar absorptivity (also known as the molar absorption coefficient or molar extinction coefficient) of a substance is a measure of how strongly it absorbs light at a particular wavelength. It is denoted by the symbol ε (epsilon) and has units of L·mol⁻¹·cm⁻¹. In this case, the given dye has a molar absorptivity of 3.85 x 10^5 L·mol⁻¹·cm⁻¹ at 585 nm.
To determine the absorbance, you can use Beer's law, which relates the absorbance of a solution to its concentration and the path length of the sample. Beer's law is expressed as:
A = εcl
where A is the absorbance, ε is the molar absorptivity, c is the concentration of the solution in moles per liter (M), and l is the path length of the sample in centimeters.
In this case, the given concentration of the dye solution is 2.50 x 10^-5 M (Molar). To find the absorbance at 585 nm, you can plug in the values into Beer's law:
A = (3.85 x 10^5 L·mol⁻¹·cm⁻¹) * (2.50 x 10^-5 M) * l
The path length, denoted by l, is usually provided in the question. However, if it is not given, you can assume a standard path length of 1 cm for most spectrophotometers.
So, if you assume a path length of 1 cm, the calculation would be:
A = (3.85 x 10^5 L·mol⁻¹·cm⁻¹) * (2.50 x 10^-5 M) * (1 cm)
= 9.625
Therefore, Simon would expect to measure an absorbance of approximately 9.625 at 585 nm for the given dye solution.
Regarding the second part of the question, drawing the visible range spectrum of the dye refers to representing the absorbance of the dye as a function of the wavelength in the visible light range (approximately 400-700 nm). You can plot a graph with absorbance on the y-axis and wavelength on the x-axis. The shape of the spectrum will depend on the dye's specific absorption properties and the range of wavelengths where it absorbs light. Without additional information about the dye, you can make a general assumption that it might exhibit a peak absorbance in the visible light range.