(a)
Interpretation:
The minimum weight of the sample for relative error due to 0.5 mg loss to be less than 0.1% needs to be determined.
Concept introduction:
The minimum weight is calculated by the relative error formula which is given below-
Here,
(b)
Interpretation:
The minimum weight of the sample for relative error due to 0.5 mg loss to be less than 0.5 % needs to be determined.
Concept introduction:
The minimum weight is calculated by the relative error formula which is given below-
Here,
(c)
Interpretation:
The minimum weight of the sample for relative error due to 0.5 mg loss to be less than 0.8 % needs to be determined.
Concept introduction:
The minimum weight is calculated by the relative error formula which is given below-
Here,
(d)
Interpretation:
The minimum weight of the sample for relative error due to 0.5 mg loss to be less than 1.2 % needs to be determined.
Concept introduction:
The minimum weight is calculated by the relative error formula which is given below-
Here,
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Principles of Instrumental Analysis
- A titrimetric method for the determination of calcium in limestone was tested with the analysis of a NIST limestone containing 30.15% CaO. The mean of the four analyzes is 30.26% CaO with a standard deviation of 0.085%. From the data accumulated from many analyzes, s→ϭ=0.094% CaO was found.a) Do the data indicate the presence of systematic error at the 95% confidence level?b) When a value for ϭ is unknown, do the data show a systematic error at the 95% confidence level?arrow_forward(a) For use in an iodine titration, you prepare a solution from 0.222 2 (+0.000 2) g of KIO3 [FM 214.001 0 (+0.000 9)] in 50.00 (+0.05) mL. Find the molarity and its uncertainty with an appropriate number of significant figures. (b) Would your answer be affected significantly if the reagent were only 99.9% pure?arrow_forwardA volumetric calcium analysis on triplicate samples of the blood serum of a patient believed to be suffering from a hyperparathyroid condition produced the following data: mmol Ca/L = 3.55, 3.65, 3.14. What is the 95% confidence interval for the mean of the data, assuming(a) No prior information about the precision of the analysis?(b) s → σ = 0.056 mmol Ca/L?arrow_forward
- 1. (a) A solution of NaCl is prepared by dissolving 0.0056 g NaCl with enough H2O in a volumetric flask to yield a final volume of 10 mL. Assuming a standard (0.1 mg) analytical balance and class A glassware is employed, calculate the concentration of the solution (M) and its corresponding error. The atomic weights of Na and Cl are 22.98976928(2) and 34.45(1), respectively (the number in parentheses indicating the error in the last digit). (b) What would the error be if a 1mL transfer pipet was employed (10x) rather than the 10 mL volumetric flask?arrow_forward(2) In the Analytical Chemistry Laboratory Mr.Yahiya Al Moqbali wanted to prepare KCI solution. He has dissolved 2.634 (+0.002) g of KCI in a 250.00 (±0.08) mL volumetric flask. Calculate the molarity of the solution and express your answer with its absolute uncertainty and give reasonable number of significant figure. [The formula mass of KCI is 74.5513 (±0.001) amu] (Show your Calculation)arrow_forward[References) A method for the determination of the corticosteroid methylprednisolone acetate in solutions obtained from pharmaceutical preparations yielded a mean value of 3.7 mg mL with a standard deviation of 0.3 mg mL. For quality control purposes, the relative uncertainty in the concentration should be no more than 3%. How many samples of each batch should be analyzed to ensure that the relative standard deviation does not exceed 9% at the 95% confidence level? samplesarrow_forward
- A volume of 250 ml of a 0.05 M solution of a reagent of formula weight (relative molecular mass) 40 was made up, the weighing being done by difference. The standard deviation of each weighing was 0.0001 g. 1 (a) What were the standard deviation and relative standard deviation of the weight of reagent used?arrow_forwardEstimate the absolute standard deviation (or uncertainty) for the results of the following calculations. Round each result so that it contains only significant digits. The numbers in parentheses are absolute standard deviations. Show your solution and express the final answer and its corresponding uncertainty. a) y= 2.998(±0.002) - 3.98 (±0.15) + 9.035 (±0.002) = 8.0531 b) y= 39.2(±0.3) x 3.054 (±0.022) x 10^ -2 = 1.197 c) y= [198(±6) - 89 (±3)] / 1335 (±2) + 64 (±7)] = 7.791 x 10^-2arrow_forwardA powder was prepared containing 3.00% NaCN and 97.00% NaCl. A sample obtained from that mixture containing 7.374×10^5 particles weighs 10.0 g. Determine the number and percent relative standard deviation of NaCN particles from a sample of the mixture weighing 6.30 g. I need nNaCN particles and %RSDarrow_forward
- A loss of 0.35 mg of Zn occurs in the course of analysis to determine the element. Calculate the percent relative error due to this loss if the masses of Zn in the replicate samples are the ff: (a) 40.0 mg (b) 400.0 mg (c) 175.0 mg (d) 600.0 mg.arrow_forwardAn excess of 0.34 mg Pb is obtained from an assay test determined in an analysis. What is the percent relative error in ppt if the mass of the sample is 550 g?arrow_forwardIn replicate analyses, the carbohydrate content of a glycoprotein (a protein with sugars attached to it) is found to be 12.6, 11.9, 13.0, 12.7, and 12.5 g of carbohydrate per 100 g of protein. (a) Find the mean of the measurements. (b) Find the standard deviation (s). (c) Find the 90% confidence intervals for the carbohydrate content. (d) If the mean and s are unchanged, but there are 10 measurements (N=10) instead of 5, what would be the confidence interval? (e) How does increasing the number of replicate measurements affect the reliability of measurements? Explain your answer. (Hint: better precision gives smaller confidence intervals)arrow_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning