the question is in three parts, i've answered the first part but i don't get the difference between the second part of the question and the third part of the question:
A standard of Pb2+ for a gunshot residue analysis using ICP-OES is
prepared by first dissolving 1.0390 g dried Pb(NO3)2 in ultrapure water
containing 1% nitric acid. The solution is brought to volume in a class A 500 mL volumetric flask with an uncertainty of ± 0.20 mL. This solution is diluted 1/10 by taking 10 mL with an Eppendorf pipette, tolerance ± 1.3 µL) and diluting this in 1% nitric acid to a final volume of 100 mL in a volumetric flask with a tolerance of ± 0.08 mL. The balance used for weighing the Pb(NO3)2 has an uncertainty of ± 0.0002 g. The atomic weights are as follows: Pb = 207.2; N = 14.01; O = 16.00.
(i) Calculate the concentration of the final solution in mg L-1 of Pb2+
and report the answer with an appropriate number of digits. [20%]
= Mr of Pb(NO3)2 = 207.2 +(14.01x2)+(16x6)
= 331.2
mole of Pb(NO3)2 = mass/Mr
= 1.0390g/(331.2)
= 3.137077295x10-3 mol
= 3.137x10-3mol
mass(in mg) = 3.137x10-3 mol x 207.2 (x1000)
absolute uncertainty of mass of Pb2+ = 650.0 mg±0.2
relative uncertainty of mass = 650.0mg±0.0308
concentration(diluted to 5.00x10-1L) = 650.0 mg/5.00x10-1 L
= 1300 mg/L
mass Pb2+ in 1.0x10-4 L = 1300mg/L x 1.0x10-4 L
= 13mg
concentration(diluted to 1.00x10-1 L) = 13mg/1.00x10-3L
130 mg/L
(ii) Determine the absolute and relative uncertainties for each of the
stages of the standard preparation. Select the largest and use this to
calculate the final Pb2+ concentration range of the standard in
mg L-1. [20%]
i don't get this question, how is it different from the next question?
iii) Calculate the concentration range in mg L-1 for the Pb2+ standard
using the proper error propagation rules. [20%]
mass(in mg) = 3.137x10-3 mol x 207.2 (x1000)
absolute uncertainty of mass of Pb2+ = 650.0 mg±0.2
relative uncertainty of mass = 650.0mg±0.0308%
absolute uncertainty of 5.00x10-1L flask = 5.00x10-1L±0.002
relative uncertainty of flask = 5.00x10-1L±0.04%
concentration(diluted to 5.00x10-1L) = 650.0 mg/5.00x10-1 L
relative uncertainty f concentration = 1300 mg/L ±0.0708%
absolute uncertainty= 1300mg/L±0.9204
absolute uncertainty of pipette= 1.0X10-4 L±1.3X10-6
Relative uncertainty= 1.0X10-4 L±1.3%
mass Pb2+ in 1.0x10-4 L = 1300mg/L x 1.0x10-4 L
relative uncertainty = 13mg±1.371%
absolute uncertainty =13mg±0.178
absolute uncertainty of flask2 = 1.00x10-1L±0.00008
relative uncertainty= 1.00x10-1L±0.08%
concentration(diluted to 1.00x10-1 L) = 13mg/1.00x10-3L
= 130 mg/L
relative uncertainty of concentration= 130 mg/L±1.45%
absolute uncertainty of concentration= 130 mg/L±1.89