EXAM-2 CHEM 2001 Summer 2006 NAME
EXAM-1 CHEM 2001 answers NAME:……………………………………….
Use the tables for statistical functions from handouts. Use the periodic table data from the table in the book. YOU MUST SHOW all your work-procedure and the execution- to receive points.
1.Convert the density D= 7.1 g / mL into pg/fL
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Answ: 7.1.g / mL
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TWO points:
2. How many grams of 90 wt % of Ca(OH)2 (FM/MM=74.093) should be diluted to the total volume of 0.250 L to make 0.2M OH- solution?
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.answ: 2.055
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3.What is and why do you need the fourth and fifth steps in chemical analysis? Any of them uses statistics?
Answ: Sample preparation
Analysis
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4. What do you do with systematic error??
Eliminate it , do not report…………………………
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5.What is the meaning of the confidence interval at confidence level of 95%? Draw an illustration of the concept.
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Answ: 95% of measurements will have values inside the range x(average +-CI
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TWO points:
6. a)For the numbers 315.9, 297.8, 316.1, 308.7, and 308.4, find the mean, standard deviation, and for CL= 95% confidence interval for the mean.
b) Is there an outliner among the data in 6a) . If there is take that into account
outliner:…………………………………….. NO OUTLINERS
X= 309.38 s= 7.469
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+- CI = …9.273……………….
TWO points
7. You have diluted 0.137+-2.1% moles of NaCl in 0.510+-0.09% Lof water. What is the molarity M ofnaCl in that solution?
Write the answer in
(a) % relative uncertainty/error and
(b) absolute uncertainty/errorwith the correct number of significant figures (“real rule”). Show the mean value +- uncertainty.
SHOW all your work
a)0.2686 +- 2.1%
b)0.2686+-0.0056
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TWO points
8. Is the probability that in one measurement you will obtain the value = 46.00g in the range of 95% or in the range of 5% , if the normal distribution established from a set of n= 8 (eight) previous measurements has the mean value = 46.12g and the standard deviation was s = 0.03g ?
a)95%
b)5%
SHOW all your work
.tcalc 11.3 vs ttab= 2.365
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YES probability is 95% NO probability is 5%
TWO POINTS
9.)Two machines for fillingbeer cans with Guinness beer were tested to find if the beer content by mass was significantly different.
Machine 1: average filled bottle mass: 100.5 +- 0.6 g, from 5measurements
Machine 2: average filled bottle mass: 101.4 +- 0.2 g, from 6 measurements
Assume that the mass of can was subtracted and does not play significant role. Make sure you SHOW what you do!!
Tcalc …3.5….. ………….. (Tab for dof)………2.262………………………….
t calculated……
Yes, there is a significant difference No, there is not
(CIRCLE ONE)
Two points
10. The mass of Pt- containing nanoclusters produced by the new method was tested against the average value of mass obtained from a well established method. These standard nanocrystals are known to have average mass of 136.8 attograms, as measured by that established method. The accompanying results were obtained:
from 7(seven) measurements, X =129.8attograms s=7.9attograms
Establish whether at the 95% C.L. the two methods are significantly different
2.344 vs 2.447
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Yes, there is a significant difference No, there is not
11.- (TWO POINTS) A linear calibration function was determined from a set of 5diluted solutions with different standard-analyte concentrations and one blank solution. The calibration function has a slope 0.9887 and intercept 0.0033. What is the concentration of the analyte in a diluted sampleif the signal from a diluted aliquote is 0.73 units and the signal of blank was 0.09 units.
Calculate the original concentration of the analyte in sample if the sample dilution was done so that 10mL of the original sample was diluted to the mark in a 250mL volumetric flask before taking it into the instrument.
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Diluted concentration…………0.643977…………………………….
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Original concentration ……………0.643977 * 250/10 = 16.1………………………….
One point
12. A solution with 100ppm of Pb is used as a stock solution to prepare standards for instrument calibration. The following volumes of that solution are taken and transferred in four separate 100 mL volumetric flasks:
5mL, 10mL, 15mL and 20mL of the stock solution are diluted separately with water to the mark in those four 100 mL flasks: flask 1, flask 2, flask 3, flask 4, respectively.
The analyte concentration is estimated to be around 30ppm. What you need to do, if anything, to correctly measure that analyte on this calibrated instrument and why? Draw the calibration function.
.DILUTE.
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Two points
13.) The concentration of an analyte A in a sample is measured on an instrument which produces linear response signal S. Two 20mL aliquots of the aqueous solution of A are transferred into two different 100mL volumetric flasks. Also, 5mL of 100ppm of the analyte A standard solution was added to the second flask. The contents of both flasks are then diluted to 100mL with water.
The signal for the solution in the first flask is S= 0.20 and in the second flask is S = 0.50. What is the concentration of the analyte A in the original sample?
SHOW YOUR WORK!
.16.666
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One point
14)What is BLANK?
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All except analyte
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15)(TWO POINTS)
The reason we use standard addition is:
Used when we have interference by matrix in the sample
The reason we use internal standard is :……
When signal oscillates
16) An analysis is performed on a sample of vine that contains unknown concentration of alcohol. A solution containing 0.0450M of ethyl alcohol (the ANALYTE) gives peak area of Aa=630 and a solution containing 0.030 M of propyl alcohol gives a peak of As =150 area on the GC instrument used for the analysis. If 10mL of 0.1M of propyl alcohol was added to the 10mL of ethyl alcohol sample and then the mixture was diluted to 30mL in a volumetric flask, the peak areas are Aa=610 and for As=710.
What is the concentration of ethyl alcohol?,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
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F=2.8
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[x] = 0.03068
17) . Which of the following is TRUE
a)An analytical lab often uses electronic balances with accuracy of 0.0001g
b)Systematic error will decrease if you increase the number of measurements (n) using the same method
c) In every day lab measurements we try to make as many measurements as it is any approximation
d) The conversion factor uncertainty/error in the expression: 1nm = 10-9m is 10-9m
e) All are NOT TRUE
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