Find the percent #"P"_2"O"_5# by mass in the fertilizer?

#"0.5000 g"# fertilizer (of some unknown #%"P"_2"O"_5# ) is dissolved in water to form #"1L"# of solution A.

A #"1.00mL"# aliquot of A is diluted to #"100 mL"# , together with a developing agent that allows the #"P"_2"O"_5# in the solution to be detected to form solution B.

Solution B is analyzed to find its absorbance at #"600 nm"# to be #0.000, 0.325, 0.553# at concentrations of #0.0, 1.2, 2.0# #"mg/L"# .

#"0.5000 g"# fertilizer (of some unknown#%"P"_2"O"_5# ) is dissolved in water to form#"1L"# of solution A. 
A
#"1.00mL"# aliquot of A is diluted to#"100 mL"# , together with a developing agent that allows the#"P"_2"O"_5# in the solution to be detected to form solution B. 
Solution B is analyzed to find its absorbance at
#"600 nm"# to be#0.000, 0.325, 0.553# at concentrations of#0.0, 1.2, 2.0# #"mg/L"# .
1 Answer
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An overview of the scenario is as follows:

#"0.5000 g"# fertilizer (of some unknown#%"P"_2"O"_5# ) is dissolved in water to form a#"1.00000 L"# #(pm "0.02 mL"# #)# solution (presumably in a volumetric flask, which is why I assume 5 decimal places).
We call this solution A. 
A
#"1.00mL"# aliquot of A (presumably via volumetric pipette) is diluted to#"100.00 mL"# (also presumably in a#100.00# #[pm "0.02 mL"# #]# volumetric flask), together with a developing agent that allows specifically the#"P"_2"O"_5# in the solution to be detected via UVVIS Spectroscopy.
We call this solution B. 
Only solution B is analyzed, and it has a known absorbance.
Thus, our calculations shall follow these steps:
 Obtain the slope (the extinction coefficient, multiplied by the path length, the result in units of
#"L/mg"# ).  Use it to calculate the concentration in
#"mg/L"# from the absorbance graph of solution B.  From the
#"P"_2"O"_5# concentration of B, get the#"P"_2"O"_5# concentration of A.  From the
#"P"_2"O"_5# concentration of A, get the mass of A that is actually#"P"_2"O"_5# .  From the experimental
#"P"_2"O"_5# mass, get the mass percent of#"P"_2"O"_5# in the original fertilizer.
OBTAINING THE SLOPE
From Beer's Law, we have
#A = epsilonbc# ,where
#epsilon# is the molar absorptivity (extinction coefficient),#b# is the path length of the cuvette (usually#"1 cm"# ), and#c# is the concentration of the analyte in#"mg/L"# .
If we plot the first three absorbances as
If you aren't familiar with Excel, here is a resource for you to look at.
GETTING THE CONCENTRATION OF SOLN B
The slope was acquired (from the LINEST function), so we can get the concentration for solution B:
#color(green)(c_B) = A/(epsilonb) = (A_B + "yint")/"slope"#
#= (0.214 + 0.001_(789))/(0.276_(05) "L/mg")#
#=# #0.781_(7)# #"mg/L"# #=># #color(green)("0.781 mg/L")# ,
where subscripts indicate the digits past the last significant digit.
GETTING THE CONCENTRATION OF SOLN A
Next, we can get the concentration of solution
#c_AV_A = c_BV_B#
#color(green)(c_A) = (c_BV_B)/(V_A)#
#= ("0.781 mg/L")("100.00 mL"/"1.00 mL")#
#=# #color(green)("78.1 mg/L")#
But, recall that the
MASS OF
Thus, to find the mass of
#color(green)(m_("P"_2"O"_5)) = "78.1 mg/L" xx "1.00000 L" = "78.1 mg P"_2"O"_5#
#= color(green)("0.0781 g P"_2"O"_5)#
MASS PERCENT OF
Therefore, the
#color(blue)(%"P"_2"O"_5)# #color(blue)("w/w") = m_("P"_2"O"_5)/(m_"fertilizer")xx100%#
#= ("0.0781 g P"_2"O"_5)/"0.5000 g fertilizer used"xx100%#
#~~ color(blue)(15.6_(3)%)#