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.

The mass of the mesh particle that reduces the \({\rm{KCl}}\)to a \(1\% \)uncertainty must be computed.

In sampling n number of particles, the standard deviation is given as

\({S_{\rm{n}}} = \sqrt {{\rm{npq}}} \)

Where,

The chance of drawing B type particles is\({\rm{q}}\)

Solution:

The mass of the mesh particle that reduces the KCl to a\({\rm{1\% }}{\rm{.}}\)uncertainty.

\(1\% \)of\({\rm{KCl}}\)particle has the same standard deviation as \(1\% \) of np.

\({\sigma _{\rm{n}}} = \sqrt {{\rm{npq}}} \)

\({\rm{p}} = 0.01{\rm{q}} = 0.99\)we get \({\rm{n}} = 9.9 \times {10^5}particles\)

Using conventional sieves, the particle diameter is estimated.

Suitable for 170/200 mesh

\( = \frac{{0.090\;{\rm{mm}}(170{\rm{ sievenumber }}) + 0.075\;{\rm{mm}}(200{\rm{ sievenumber }})}}{2} = 0.0825\;{\rm{mm}}\)

\(\frac{4}{3}\pi {(0.0825\;{\rm{mm}})^3} = 2.35 \times {10^{ - 3}}\;{\rm{mL}}\)is used to compute the particle volume.

The mass of 1% chloride is computed as follows:

\(Mass = \left( {9.9 \times {{10}^5}} \right.particle)\left( {0.00236 \times \times {{10}^{ - 6}}\;{\rm{mL}}/} \right.particle)(2.108\;{\rm{g}}/{\rm{mL}}) = 0.61\;{\rm{g}}\)

Conclusion

The mass of the mesh particle that reduces the \({\rm{KCl}}\)to a\({\rm{1\% }}\)uncertainty was computed.