Question: Consider a random mixture containing \(4.00\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) (density\(2.532g/mL\)) and \(96.00\;{\rm{g}}\)of \({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\) (density\(2.428\;{\rm{g}}/{\rm{mL}}\)) with a uniform spherical particle radius of\(0.075\;{\rm{mm}}\).

(a) Calculate the mass of a single particle of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) and the number of particles of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) in the mixture. Do the same for\({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\).

(b) What is the expected number of particles in \(0.100\;{\rm{g}}\)of the mixture?

(c) Calculate the relative sampling standard deviation in the number of particles of each type in a \(0.100\;{\rm{g}}\)sample of the mixture.

• The standard deviation is a measure of how far something deviates from the mean (for example, spread, dispersion,). 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.

a)

First we will calculate the volume with the following:

\(\begin{array}{l}V = \frac{4}{3}\pi {r^3}\\V = \frac{4}{3}\pi {\left( {7.5 \times {{10}^{ - 3}}\;{\rm{cm}}} \right)^3}\\V = 1.767 \times {10^{ - 6}}\;{\rm{mL}}\end{array}\)

Next calculate the mass of particles in\({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\)and \({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\) :

The equation used is\(m = \frac{V}{\rho }\)

(1) \(m\left( {{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}} \right) = \frac{{1.767 \times {{10}^{ - 6}}\;{\rm{mL}}}}{{2.532\;{\rm{g}}/{\rm{mL}}}} = 4.474 \times {10^{ - 6}}\;{\rm{g}}\)

(2)\(m\left( {\;{{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}} \right) = \frac{{1.767 \times {{10}^{ - 6}}\;{\rm{mL}}}}{{2.428\;{\rm{g}}/{\rm{mL}}}} = 4.291 \times {10^{ - 6}}\;{\rm{g}}\)

Then we can calculate the number of particles for each compound

The equation used is\({n_{{\rm{particles }}}} = \frac{{{m_{{\rm{compound }}}}}}{{{m_{{\rm{particles }}}}}}\)

(1)\({n_{{\rm{particles }},1}}\left( {{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}} \right) = \frac{{4.0\;{\rm{g}}}}{{4.474 \times {{10}^{ - 6}}\;{\rm{g}}}} = 8.941 \times {10^5}\)

(2) \({n_{{\rm{particles }},2}}\left( {\;{{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}} \right) = \frac{{96.0\;{\rm{g}}}}{{4.291 \times {{10}^{ - 6}}\;{\rm{g}}}} = 2.237 \times {10^7}\)

Also, we should calculate the fraction of each compound:

(1)\({f_1} = \frac{{{n_{{\rm{particles, }}1}}}}{{{n_{{\rm{particles }},1}} + {n_{{\rm{particles }},2}}}} = \frac{{8.941 \times {{10}^5}}}{{\left( {8.941 \times {{10}^5}} \right) + \left( {2.237 \times {{10}^7}} \right)}} = 0.0384\)

(2)\({f_2} = \frac{{{n_{{\rm{particles }},1}}}}{{{n_{{\rm{particles }},1}} + {n_{{\rm{particles }},2}}}} = \frac{{2.237 \times {{10}^7}}}{{\left( {8.941 \times {{10}^5}} \right) + \left( {2.237 \times {{10}^7}} \right)}} = 0.962\)

Therefore,

\((i)\)The mass of the particle in \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\)is \(4.474 \times {10^{ - 6}}\;{\rm{g}}\) and the number of particles is \(8.941 \times {10^5}\).

\((ii)\) The mass of the particle in \({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\)is \(4.291 \times {10^{ - 6}}\;{\rm{g}}\) and the number of particles is \(2.237 \times {10^7}\).

b)

The expected number of particles in\(0.1\;{\rm{g}}\) of the mixture would be:

\(\begin{array}{l}{n_{{\rm{particles }}({\rm{ total }})}} = {n_{{\rm{particles }},1}} + {n_{{\rm{particles }},2}}\\{n_{{\rm{particles(total }})}} = \left( {8.941 \times {{10}^5}} \right) + \left( {2.237 \times {{10}^7}} \right)\\{n_{{\rm{particles(total }})}} = 2.326 \times {10^7}{\rm{ in }}100\;{\rm{g}}\end{array}\)

\(\begin{array}{l}{n_{{\rm{particles(expected }})}} = \frac{{{n_{{\rm{particles }}({\rm{ total }})}}}}{{{{10}^3}}}\\{n_{{\rm{particles }}({\rm{ expected }})}} = \frac{{2.326 \times {{10}^7}}}{{{{10}^3}}}\\{n_{{\rm{particles(expected }})}} = 2.326 \times {10^4}{\rm{ in }}0.1\;{\rm{g}}\end{array}\)

Note that\({10^3}\)is written because \(100/0.1 = 1000 = {10^3}\)

Hence, the expected number of the particle is \(2.326 \times {10^4}{\rm{ in }}0.1\;{\rm{g}}\)

c)

Here note that expected number of each particle in\(0.1\;{\rm{g}}\) is \({10^{ - 3}}\) of \({n_{{\rm{particles }},1}}\) and \({n_{{\rm{particles }},2}}\)so the following is:

(1)\({n_{{\rm{particles, }}1}}\left( {{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}} \right) = 8.941 \times {10^2}\)

(2)\({n_{{\rm{particles, }}2}}\left( {\;{{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}} \right) = 2.237 \times {10^4}\)

First determine the value of sampling standard deviation:

Sampling standard deviation\( = \sqrt {npq} \)

Sampling standard deviation\( = \sqrt {\left( {2.326 \times {{10}^4}} \right)(0.0384)(0.962)} \)

Sampling standard deviation \( = 29.3\)

Next calculate the relative sampling standard deviations:

\({\rm{ (1) RSTD for}}\,{\rm{N}}{{\rm{a}}_2}C{O_3} = \frac{{29.3}}{{8.941 \times {{10}^2}}} = 3.28\% \)

(2) RSTD for\({{\rm{K}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\)\( = \frac{{29.3}}{{2.237 \times {{10}^4}}} = 0.131\% \)

Therefore, the relative sampling standard deviation is\(3.28\% \)and \(0.131\% \).