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What mass of sample in Figure 28-3 is expected to give a sampling standard deviation of \( \pm 6\% \)?

Short Answer

Expert verified

The mass of the sample which produces the relative standard deviation of \(1\% \)

Step by step solution

01

Definition of Standard deviation.

  • The standard deviation is a measure of how far something deviates from the mean (for example, spread, dispersion, or spread). A "typical" variation from the mean is represented by the standard deviation.
  • Because it returns to the data set's original units of measurement, it's a common measure of variability.
  • The standard deviation, defined as the square root of the variance, is a statistic that represents the dispersion of a dataset relative to its mean.
02

Determine the Sampling Standard deviation.

In this task we will calculate the mass of sample in Figure \(28 - 3\) expected to give a sampling standard deviation of\( \pm 6\% \).

Here we will use the equation \(28 - 5\) for relative variance\(\left( {{R^2}} \right)\) :

\(\begin{array}{l}{R^2} = {\left( {\frac{{{\delta _n}}}{n}} \right)^2}\\ = \frac{{pq}}{n}n\\{R^2} = pq\end{array}\)

Where, \(p\) and \(q\) are the fraction of particles present

Next we will rearrange the equation \(28 - 5\)to the following and calculate the mass of sample:

\(\begin{array}{l}m{R^2} = {K_a}\\m = \frac{{{K_a}}}{{{R^2}}}\\m = \frac{{36\;{\rm{g}}}}{{{6^2}}}\\m = 1\;{\rm{g}}\end{array}\)

Where, \(R\) is the relative standard deviation and \({K_a}\) is the sampling constant

Therefore, the mass of the sample which produces the relative standard deviation of \(1\% \)

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Most popular questions from this chapter

By what factor must the mass increase to reduce the sampling standard deviation by a factor of 2?

Acid-base equilibria of Cr(III) were summarized in Problem 10-36. Cr(VI) in aqueous solution above pH 6 exists as the yellow tetrahedral chromate ion, \({\rm{CrO}}_4^{2 - }.\)Between\({\rm{pH}}2\)and \(6,{\rm{Cr}}\)(VI) exists as an equilibrium mixture of\({\rm{HCrO}}_4^ - \) and orange-red dichromate,\({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }.{\rm{Cr}}({\rm{VI}})\) is a carcinogen, but \({\rm{Cr }}(III)\)is not considered to be as harmful. The following procedure was used to measure\({\rm{Cr }}({\rm{VI}})\) in airborne particulate matter in workplaces.

1. Particles were collected by drawing a known volume of air through a polyvinyl chloride filter with \(5 - \mu {\rm{M}}\)pore size.

2. The filter was placed in a centrifuge tube and \(10\;{\rm{mL}}\)of \(0.05{\rm{M}}{\left( {{\rm{N}}{{\rm{H}}_4}} \right)_2}{\rm{S}}{{\rm{O}}_4}/0.05{\rm{MN}}{{\rm{H}}_3}buffer,{\rm{pH}}8,\) were added. The immersed filter was agitated by ultrasonic vibration for\(30\;{\rm{min}}\)at \({35^\circ }{\rm{C}}\)to extract all \({\rm{Cr }}(III)and{\rm{Cr}}\)(VI) into solution.

3. A measured volume of extract was passed through a "strongly basic" anion exchanger (Table 26-1) in the \({\rm{C}}{{\rm{l}}^ - }\)form. Then the resin was washed with distilled water. Liquid containing \({\rm{Cr}}\)(III) from the extract and the wash was discarded.

4. Cr(VI) was then eluted from the column with\(0.5{\rm{M}}{\left( {{\rm{N}}{{\rm{H}}_4}} \right)_2}{\rm{S}}{{\rm{O}}_4}/0.05{\rm{MN}}{{\rm{H}}_3}\) buffer, \({\rm{pH}}8,\)and collected in a vial.

5. The eluted \({\rm{Cr}}\)(VI) solution was acidified with \({\rm{HCl}}\)and treated with a solution of 1,5 -diphenylcarbazide, a reagent that forms a colored complex with Cr(VI). The concentration of the complex was measured by its visible absorbance.

(a) What are the dominant species of \({\rm{Cr}}\)(VI) and \({\rm{Cr}}\)(III) at\({\rm{pH}}8\)?

(b) What is the purpose of the anion exchanger in step 3 ?

(c) Why is a "strongly basic" anion exchanger used instead of a "weakly basic" exchanger?

(d) Why is Cr(VI) eluted in step 4 but not step 3 ?

EXAMPLE- Particles designated \(50/00\)mesh pass through a 50 mesh sieve bou are retained by a lo0 mesh sieve. Their size is in the range 0.150-0.300 mm.

does not pass is retained for your sample. This procedure gives particles whose diameters are in the range \(0.85 - 1.18\;{\rm{mm}}.\) We refer to the size range as \(16/20{\rm{mesh}}.\)

Suppose that much finer particles of \(80/120\)mesh size (average diameter \( = 152\mu {\rm{m}},\) average volume\( = 1.84\;{\rm{nL}}\)) were used instead. Now the mass containing \({10^4}\) particles is reduced from \(11.0to0.0388\;{\rm{g}}.\) We could analyze a larger sample to reduce the sampling uncertainty for chloride.

Why is it advantageous to use large particles \(\left( {{\bf{50}}{\rm{ }}\mu {\bf{m}}} \right)\) for solid phase extraction, but small particles \(\left( {{\bf{5}}{\rm{ }}\mu {\bf{m}}} \right)\) for chromatography?

An example of a mixture of 1-mm-diameter particles of \({\rm{KCl}}\)and \({\rm{KN}}{{\rm{O}}_3}\)in a number ratio \(1:99\)follows Equation 28-4. A sample containing \({10^4}\)particles weighs\(11.0\;{\rm{g}}\). What is the expected number and relative standard deviation of \({\rm{KCl}}\)particles in a sample weighing\(11.0 \times {10^2}\;{\rm{g}}\)?

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