Question

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\(11.0 \times {10^2}\;{\rm{g}}\)?

Short answer

The number of \({\rm{KCl}}\) particles is \({10^4}\) and the relative standard deviation of the \({\rm{KCl}}\) particles is\(99\% \).

Step-by-step solution

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.

Determine the Standard deviation

First determine the number of particles in \(11 \times {10^2}\;{\rm{g}}\) considering that we know that \(11\;{\rm{g}}\) contains \({10^4}\) particles:

\(11\;{\rm{g}} = {10^4}\)particles

\(\begin{aligned}{}11 \times {10^2} &= \left( {{{10}^4} \times {{10}^2}} \right){\rm{ particles }}\\11 \times {10^2} &= {10^6}{\rm{ particles}}\end{aligned}\)

Next we will calculate the number of \({\rm{KCl}}\) particles:

\(\begin{aligned}{}n({\rm{KCl}}) &= n \times p\\n({\rm{KCl}}) &= {10^6} \times 0.01\\n({\rm{KCl}}) &= {10^4}\end{aligned}\)

Then we will calculate the relative standard deviation by the following equation

\({\rm{ Relative standard deviation }} = \sqrt {npq} /n({\rm{KCl}})\)

Relative standard deviation \( = \sqrt {{{10}^6} \times (0.01) \times 0.99} /{10^4}\)

Relative standard deviation \( = 99\% \)

Therefore, the relative standard deviation of the \({\rm{KCl}}\) particles is\(99\% \).