Any help/advice on answering this question would be a big help! :)

The analysis of lead (Pb) in a sample of solder was carried out using atomic absorption spectroscopy as follows.
1.013 g of solder was completely dissolved in a small volume of nitric acid and the solution made up to 1.000 L with distilled water. A 50.0 μL aliquot of this was then diluted to 100.0 mL with distilled water, and when analysed by atomic absorption in an air/acetylene flame it gave an absorbance of 0.215 at 283.3 nm. The instrument was calibrated, after zeroing with distilled water, with standard solutions of lead containing 0.100 and 0.300 ppm Pb which gave absorbances of 0.120 and 0.360 respectively.
Calculate the wt % of Pb in the solder.

I would proceed this way.

A = abc and since these measurements were made with the same length cell (I assume) we an ignore that. We need to
1. graph the data for a working curve (A vs C) or
2. determine a from the data and that is easily done.
A = a*c
0.120 = a*0.1 and
0.360 = a*0.3. Solv for a and in both cases a = 1.2. That allows to you determine c for the 100 mL sample.
A = a*c; 0.215 = 1.2*c and calculate c in ppm. I'll estimate that at about 0.18 ppm or so. That's c in the 100 mL. What was it in the 1.00 L? That's approx
0.18 x 100/0.05 = about 360 ppm
You know 1 ppm is 1 g Pb in 1o^6 mL solution.
Convert 360 ppm to g in the 1000 mL, then
%Pb = (g Pb in sample/g sample)*100 = ?
Post your work if you get stuck.

DrBob222, thank you for the help!

To calculate the weight percent of Pb in the solder, we need to follow a series of steps:

Step 1: Calculate the concentration of Pb in the 50.0 μL aliquot of the sample.
We know that an absorbance reading of 0.215 was obtained for the 50.0 μL aliquot of the sample. First, we need to find the concentration of Pb in the aliquot using the calibration curve of the instrument.

Using the calibration data provided, we can see that an absorbance reading of 0.120 corresponds to a concentration of 0.100 ppm, and an absorbance of 0.360 corresponds to a concentration of 0.300 ppm.

To calculate the concentration of Pb in the aliquot, we can set up a proportion:

(0.215 - 0.120) / (0.360 - 0.120) = (x - 0.100) / (0.300 - 0.100)

Solving for x, the concentration of Pb in the aliquot, we find:

(0.095) / (0.240) = (x - 0.100) / (0.200)

x - 0.100 = (0.095 / 0.240) * 0.200

x - 0.100 = 0.0792

x = 0.1792 ppm

Step 2: Calculate the concentration of Pb in the original 1.000 L solution.
We need to account for the dilution of the aliquot. Since it was diluted to 100.0 mL, the original 1.000 L solution was diluted by a factor of 1000/100 = 10.

Therefore, the concentration of Pb in the original solution is:

(0.1792 ppm) * 10 = 1.792 ppm

Step 3: Calculate the amount of Pb in the original solution in grams.
To do this, we need to know the molar mass of Pb. The molar mass of Pb is 207.2 g/mol.

First, calculate the number of moles of Pb in the original solution:

(1.792 ppm / 1,000,000 ppm) * 1.000 L = 0.001792 mol

Then, calculate the mass of Pb in grams:

0.001792 mol * 207.2 g/mol = 0.3717 g

Step 4: Calculate the weight percent of Pb in the solder.
The weight percent of Pb is given by:

(0.3717 g / 1.013 g) * 100% = 36.6%

Therefore, the weight percent of Pb in the solder sample is approximately 36.6%.