91Ó°ÊÓ

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.