Chapter 3 0f Quantitative Chemical Analysis 9th Edition - Question 22 Express th
ID: 536922 • Letter: C
Question
Chapter 3 0f Quantitative Chemical Analysis 9th Edition - Question 22
Express the molecular mass (plusminus uncertainty) of C_ H_ O_6N_ 3, with the correct number of significant figures. You have a stock solution certified by a manufacturer to contain 150.0 plusminus 0.3 mu g SO^2-_4/mL. You would to dilute it by a factor of 100 to obtain 500 mu g/mL. Two possible methods of dilution are stated below. For each method, calculate the resulting uncertainty in concentration. Use manufacturer's tolerances in Tables 2-3 and 2-4 for uncertainties. Explain why one method is more precise than the other. (a) Dilute 10.00 mL up to 100 mL with a transfer pipet and volumetric flask. Then take 10.00 mL of the dilute solution and dilute it again to 100 mL. (b) Dilute 1.000 mL up to 100 mL with a transfer pipet and volumetric flask. Avogadro's number can be computed from the following measured properties of pure crystalline silicon: ' (1) atomic massExplanation / Answer
For 100mL flask tolerance is ±0.08 mL
For 10mL transfer pipets tolerance is ± 0.02 mL
Concentration after 1st dilution = [(150±0.3)g/mL* (10±0.02)mL] / (100±0.08)mL
[150 *(10/100)] ± [150 *(10/100)] * [ (0.3/150) + (0.02/10) + (0.08/100) ]
= (15 ± 0.072) g/mL
After 2nd dilution the concentration will be,
[ (15 ± 0.072) g/mL*(10 ± 0.02) mL ] / (100 ± 0.08) mL
[15*(10/100)] ± [15*(10/100)] * [ (0.072/15) + (0.02/10) + (0.08/100) ]
= (1.5 ± 0.0114) g/mL
b) For 1mL transfer pipet tolerance is ± 0.006 mL
Concentration = [(150 ± 0.3) g/mL*(1 ± 0.006) mL ] / (100 ± 0.08) mL
[150*(1/100) ] ± [150*(1/100) ] * [(0.3/150) + (0.006/1) + (0.08/100) ]
= (1.5 ± 0.0132) g/mL
If you compare both the errors you can observe that a) has less error than b)
So the first method a) gives seems to be more precise
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.